Precalculus Examples

$| 3 x^{2} - 7 x + 2 | > 100$

Step 1

Write $| 3 x^{2} - 7 x + 2 | > 100$ as a piecewise.

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Step 1.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$3 x^{2} - 7 x + 2 \geq 0$

Step 1.2

Solve the inequality.

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Step 1.2.1

Convert the inequality to an equation.

$3 x^{2} - 7 x + 2 = 0$

Step 1.2.2

Factor by grouping.

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Step 1.2.2.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot 2 = 6$ and whose sum is $b = - 7$ .

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Step 1.2.2.1.1

Factor $- 7$ out of $- 7 x$ .

$3 x^{2} - 7 x + 2 = 0$

Step 1.2.2.1.2

Rewrite $- 7$ as $- 1$ plus $- 6$

$3 x^{2} + (- 1 - 6) x + 2 = 0$

Step 1.2.2.1.3

Apply the distributive property.

$3 x^{2} - 1 x - 6 x + 2 = 0$

Step 1.2.2.2

Factor out the greatest common factor from each group.

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Step 1.2.2.2.1

Group the first two terms and the last two terms.

$(3 x^{2} - 1 x) - 6 x + 2 = 0$

Step 1.2.2.2.2

Factor out the greatest common factor (GCF) from each group.

$x (3 x - 1) - 2 (3 x - 1) = 0$

Step 1.2.2.3

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

$(3 x - 1) (x - 2) = 0$

Step 1.2.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$3 x - 1 = 0$

$x - 2 = 0$

Step 1.2.4

Set $3 x - 1$ equal to $0$ and solve for $x$ .

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Step 1.2.4.1

Set $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Step 1.2.4.2

Solve $3 x - 1 = 0$ for $x$ .

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Step 1.2.4.2.1

Add $1$ to both sides of the equation.

$3 x = 1$

Step 1.2.4.2.2

Divide each term in $3 x = 1$ by $3$ and simplify.

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Step 1.2.4.2.2.1

Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Step 1.2.4.2.2.2

Simplify the left side.

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Step 1.2.4.2.2.2.1

Cancel the common factor of $3$ .

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Step 1.2.4.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Step 1.2.4.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{3}$

Step 1.2.5

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 1.2.5.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 1.2.5.2

Add $2$ to both sides of the equation.

$x = 2$

Step 1.2.6

The final solution is all the values that make $(3 x - 1) (x - 2) = 0$ true.

$x = \frac{1}{3}, 2$

Step 1.2.7

Use each root to create test intervals.

$x < \frac{1}{3}$

$\frac{1}{3} < x < 2$

$x > 2$

Step 1.2.8

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 1.2.8.1

Test a value on the interval $x < \frac{1}{3}$ to see if it makes the inequality true.

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Step 1.2.8.1.1

Choose a value on the interval $x < \frac{1}{3}$ and see if this value makes the original inequality true.

$x = 0$

Step 1.2.8.1.2

Replace $x$ with $0$ in the original inequality.

$3 {(0)}^{2} - 7 \cdot 0 + 2 \geq 0$

Step 1.2.8.1.3

The left side $2$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.2.8.2

Test a value on the interval $\frac{1}{3} < x < 2$ to see if it makes the inequality true.

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Step 1.2.8.2.1

Choose a value on the interval $\frac{1}{3} < x < 2$ and see if this value makes the original inequality true.

$x = 1$

Step 1.2.8.2.2

Replace $x$ with $1$ in the original inequality.

$3 {(1)}^{2} - 7 \cdot 1 + 2 \geq 0$

Step 1.2.8.2.3

The left side $- 2$ is less than the right side $0$ , which means that the given statement is false.

False

Step 1.2.8.3

Test a value on the interval $x > 2$ to see if it makes the inequality true.

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Step 1.2.8.3.1

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

Step 1.2.8.3.2

Replace $x$ with $4$ in the original inequality.

$3 {(4)}^{2} - 7 \cdot 4 + 2 \geq 0$

Step 1.2.8.3.3

The left side $22$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.2.8.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < \frac{1}{3}$ True

$\frac{1}{3} < x < 2$ False

$x > 2$ True

$x < \frac{1}{3}$ True

$\frac{1}{3} < x < 2$ False

$x > 2$ True

Step 1.2.9

The solution consists of all of the true intervals.

$x \leq \frac{1}{3}$ or $x \geq 2$

Step 1.3

In the piece where $3 x^{2} - 7 x + 2$ is non-negative, remove the absolute value.

$3 x^{2} - 7 x + 2 > 100$

Step 1.4

To find the interval for the second piece, find where the inside of the absolute value is negative.

$3 x^{2} - 7 x + 2 < 0$

Step 1.5

Solve the inequality.

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Step 1.5.1

Convert the inequality to an equation.

$3 x^{2} - 7 x + 2 = 0$

Step 1.5.2

Factor by grouping.

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Step 1.5.2.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot 2 = 6$ and whose sum is $b = - 7$ .

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Step 1.5.2.1.1

Factor $- 7$ out of $- 7 x$ .

$3 x^{2} - 7 x + 2 = 0$

Step 1.5.2.1.2

Rewrite $- 7$ as $- 1$ plus $- 6$

$3 x^{2} + (- 1 - 6) x + 2 = 0$

Step 1.5.2.1.3

Apply the distributive property.

$3 x^{2} - 1 x - 6 x + 2 = 0$

Step 1.5.2.2

Factor out the greatest common factor from each group.

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Step 1.5.2.2.1

Group the first two terms and the last two terms.

$(3 x^{2} - 1 x) - 6 x + 2 = 0$

Step 1.5.2.2.2

Factor out the greatest common factor (GCF) from each group.

$x (3 x - 1) - 2 (3 x - 1) = 0$

Step 1.5.2.3

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

$(3 x - 1) (x - 2) = 0$

Step 1.5.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$3 x - 1 = 0$

$x - 2 = 0$

Step 1.5.4

Set $3 x - 1$ equal to $0$ and solve for $x$ .

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Step 1.5.4.1

Set $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Step 1.5.4.2

Solve $3 x - 1 = 0$ for $x$ .

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Step 1.5.4.2.1

Add $1$ to both sides of the equation.

$3 x = 1$

Step 1.5.4.2.2

Divide each term in $3 x = 1$ by $3$ and simplify.

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Step 1.5.4.2.2.1

Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Step 1.5.4.2.2.2

Simplify the left side.

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Step 1.5.4.2.2.2.1

Cancel the common factor of $3$ .

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Step 1.5.4.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Step 1.5.4.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{3}$

Step 1.5.5

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 1.5.5.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 1.5.5.2

Add $2$ to both sides of the equation.

$x = 2$

Step 1.5.6

The final solution is all the values that make $(3 x - 1) (x - 2) = 0$ true.

$x = \frac{1}{3}, 2$

Step 1.5.7

Use each root to create test intervals.

$x < \frac{1}{3}$

$\frac{1}{3} < x < 2$

$x > 2$

Step 1.5.8

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 1.5.8.1

Test a value on the interval $x < \frac{1}{3}$ to see if it makes the inequality true.

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Step 1.5.8.1.1

Choose a value on the interval $x < \frac{1}{3}$ and see if this value makes the original inequality true.

$x = 0$

Step 1.5.8.1.2

Replace $x$ with $0$ in the original inequality.

$3 {(0)}^{2} - 7 \cdot 0 + 2 < 0$

Step 1.5.8.1.3

The left side $2$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 1.5.8.2

Test a value on the interval $\frac{1}{3} < x < 2$ to see if it makes the inequality true.

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Step 1.5.8.2.1

Choose a value on the interval $\frac{1}{3} < x < 2$ and see if this value makes the original inequality true.

$x = 1$

Step 1.5.8.2.2

Replace $x$ with $1$ in the original inequality.

$3 {(1)}^{2} - 7 \cdot 1 + 2 < 0$

Step 1.5.8.2.3

The left side $- 2$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 1.5.8.3

Test a value on the interval $x > 2$ to see if it makes the inequality true.

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Step 1.5.8.3.1

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

Step 1.5.8.3.2

Replace $x$ with $4$ in the original inequality.

$3 {(4)}^{2} - 7 \cdot 4 + 2 < 0$

Step 1.5.8.3.3

The left side $22$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 1.5.8.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < \frac{1}{3}$ False

$\frac{1}{3} < x < 2$ True

$x > 2$ False

$x < \frac{1}{3}$ False

$\frac{1}{3} < x < 2$ True

$x > 2$ False

Step 1.5.9

The solution consists of all of the true intervals.

$\frac{1}{3} < x < 2$

Step 1.6

In the piece where $3 x^{2} - 7 x + 2$ is negative, remove the absolute value and multiply by $- 1$ .

$- (3 x^{2} - 7 x + 2) > 100$

Step 1.7

Write as a piecewise.

${\begin{cases} 3 x^{2} - 7 x + 2 > 100 & x \leq \frac{1}{3} or x \geq 2 \\ - (3 x^{2} - 7 x + 2) > 100 & \frac{1}{3} < x < 2 \end{cases}$

Step 1.8

Simplify $- (3 x^{2} - 7 x + 2) > 100$ .

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Step 1.8.1

Apply the distributive property.

${\begin{cases} 3 x^{2} - 7 x + 2 > 100 & x \leq \frac{1}{3} or x \geq 2 \\ - (3 x^{2}) - (- 7 x) - 1 \cdot 2 > 100 & \frac{1}{3} < x < 2 \end{cases}$

Step 1.8.2

Simplify.

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Step 1.8.2.1

Multiply $3$ by $- 1$ .

${\begin{cases} 3 x^{2} - 7 x + 2 > 100 & x \leq \frac{1}{3} or x \geq 2 \\ - 3 x^{2} - (- 7 x) - 1 \cdot 2 > 100 & \frac{1}{3} < x < 2 \end{cases}$

Step 1.8.2.2

Multiply $- 7$ by $- 1$ .

${\begin{cases} 3 x^{2} - 7 x + 2 > 100 & x \leq \frac{1}{3} or x \geq 2 \\ - 3 x^{2} + 7 x - 1 \cdot 2 > 100 & \frac{1}{3} < x < 2 \end{cases}$

Step 1.8.2.3

Multiply $- 1$ by $2$ .

${\begin{cases} 3 x^{2} - 7 x + 2 > 100 & x \leq \frac{1}{3} or x \geq 2 \\ - 3 x^{2} + 7 x - 2 > 100 & \frac{1}{3} < x < 2 \end{cases}$

Step 2

Solve $3 x^{2} - 7 x + 2 > 100$ for $x$ .

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Step 2.1

Convert the inequality to an equation.

$3 x^{2} - 7 x + 2 = 100$

Step 2.2

Subtract $100$ from both sides of the equation.

$3 x^{2} - 7 x + 2 - 100 = 0$

Step 2.3

Subtract $100$ from $2$ .

$3 x^{2} - 7 x - 98 = 0$

Step 2.4

Factor by grouping.

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Step 2.4.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot - 98 = - 294$ and whose sum is $b = - 7$ .

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Step 2.4.1.1

Factor $- 7$ out of $- 7 x$ .

$3 x^{2} - 7 x - 98 = 0$

Step 2.4.1.2

Rewrite $- 7$ as $14$ plus $- 21$

$3 x^{2} + (14 - 21) x - 98 = 0$

Step 2.4.1.3

Apply the distributive property.

$3 x^{2} + 14 x - 21 x - 98 = 0$

Step 2.4.2

Factor out the greatest common factor from each group.

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Step 2.4.2.1

Group the first two terms and the last two terms.

$(3 x^{2} + 14 x) - 21 x - 98 = 0$

Step 2.4.2.2

Factor out the greatest common factor (GCF) from each group.

$x (3 x + 14) - 7 (3 x + 14) = 0$

Step 2.4.3

Factor the polynomial by factoring out the greatest common factor, $3 x + 14$ .

$(3 x + 14) (x - 7) = 0$

Step 2.5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$3 x + 14 = 0$

$x - 7 = 0$

Step 2.6

Set $3 x + 14$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $3 x + 14$ equal to $0$ .

$3 x + 14 = 0$

Step 2.6.2

Solve $3 x + 14 = 0$ for $x$ .

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Step 2.6.2.1

Subtract $14$ from both sides of the equation.

$3 x = - 14$

Step 2.6.2.2

Divide each term in $3 x = - 14$ by $3$ and simplify.

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Step 2.6.2.2.1

Divide each term in $3 x = - 14$ by $3$ .

$\frac{3 x}{3} = \frac{- 14}{3}$

Step 2.6.2.2.2

Simplify the left side.

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Step 2.6.2.2.2.1

Cancel the common factor of $3$ .

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Step 2.6.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 14}{3}$

Step 2.6.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 14}{3}$

Step 2.6.2.2.3

Simplify the right side.

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Step 2.6.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{14}{3}$

Step 2.7

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 2.7.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 2.7.2

Add $7$ to both sides of the equation.

$x = 7$

Step 2.8

The final solution is all the values that make $(3 x + 14) (x - 7) = 0$ true.

$x = - \frac{14}{3}, 7$

Step 2.9

Use each root to create test intervals.

$x < - \frac{14}{3}$

$- \frac{14}{3} < x < 7$

$x > 7$

Step 2.10

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 2.10.1

Test a value on the interval $x < - \frac{14}{3}$ to see if it makes the inequality true.

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Step 2.10.1.1

Choose a value on the interval $x < - \frac{14}{3}$ and see if this value makes the original inequality true.

$x = - 7$

Step 2.10.1.2

Replace $x$ with $- 7$ in the original inequality.

$3 {(- 7)}^{2} - 7 \cdot - 7 + 2 > 100$

Step 2.10.1.3

The left side $198$ is greater than the right side $100$ , which means that the given statement is always true.

True

Step 2.10.2

Test a value on the interval $- \frac{14}{3} < x < 7$ to see if it makes the inequality true.

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Step 2.10.2.1

Choose a value on the interval $- \frac{14}{3} < x < 7$ and see if this value makes the original inequality true.

$x = 0$

Step 2.10.2.2

Replace $x$ with $0$ in the original inequality.

$3 {(0)}^{2} - 7 \cdot 0 + 2 > 100$

Step 2.10.2.3

The left side $2$ is not greater than the right side $100$ , which means that the given statement is false.

False

Step 2.10.3

Test a value on the interval $x > 7$ to see if it makes the inequality true.

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Step 2.10.3.1

Choose a value on the interval $x > 7$ and see if this value makes the original inequality true.

$x = 10$

Step 2.10.3.2

Replace $x$ with $10$ in the original inequality.

$3 {(10)}^{2} - 7 \cdot 10 + 2 > 100$

Step 2.10.3.3

The left side $232$ is greater than the right side $100$ , which means that the given statement is always true.

True

Step 2.10.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < - \frac{14}{3}$ True

$- \frac{14}{3} < x < 7$ False

$x > 7$ True

$x < - \frac{14}{3}$ True

$- \frac{14}{3} < x < 7$ False

$x > 7$ True

Step 2.11

The solution consists of all of the true intervals.

$x < - \frac{14}{3}$ or $x > 7$

Step 3

Solve $- 3 x^{2} + 7 x - 2 > 100$ for $x$ .

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Step 3.1

Move all terms to the left side of the equation and simplify.

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Step 3.1.1

Subtract $100$ from both sides of the inequality.

$- 3 x^{2} + 7 x - 2 - 100 > 0$

Step 3.1.2

Subtract $100$ from $- 2$ .

$- 3 x^{2} + 7 x - 102 > 0$

Step 3.2

Convert the inequality to an equation.

$- 3 x^{2} + 7 x - 102 = 0$

Step 3.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3.4

Substitute the values $a = - 3$ , $b = 7$ , and $c = - 102$ into the quadratic formula and solve for $x$ .

$\frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (- 3 \cdot - 102)}}{2 \cdot - 3}$

Step 3.5

Simplify.

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Step 3.5.1

Simplify the numerator.

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Step 3.5.1.1

Raise $7$ to the power of $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot - 3 \cdot - 102}}{2 \cdot - 3}$

Step 3.5.1.2

Multiply $- 4 \cdot - 3 \cdot - 102$ .

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Step 3.5.1.2.1

Multiply $- 4$ by $- 3$ .

$x = \frac{- 7 \pm \sqrt{49 + 12 \cdot - 102}}{2 \cdot - 3}$

Step 3.5.1.2.2

Multiply $12$ by $- 102$ .

$x = \frac{- 7 \pm \sqrt{49 - 1224}}{2 \cdot - 3}$

Step 3.5.1.3

Subtract $1224$ from $49$ .

$x = \frac{- 7 \pm \sqrt{- 1175}}{2 \cdot - 3}$

Step 3.5.1.4

Rewrite $- 1175$ as $- 1 (1175)$ .

$x = \frac{- 7 \pm \sqrt{- 1 \cdot 1175}}{2 \cdot - 3}$

Step 3.5.1.5

Rewrite $\sqrt{- 1 (1175)}$ as $\sqrt{- 1} \cdot \sqrt{1175}$ .

$x = \frac{- 7 \pm \sqrt{- 1} \cdot \sqrt{1175}}{2 \cdot - 3}$

Step 3.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 7 \pm i \cdot \sqrt{1175}}{2 \cdot - 3}$

Step 3.5.1.7

Rewrite $1175$ as $5^{2} \cdot 47$ .

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Step 3.5.1.7.1

Factor $25$ out of $1175$ .

$x = \frac{- 7 \pm i \cdot \sqrt{25 (47)}}{2 \cdot - 3}$

Step 3.5.1.7.2

Rewrite $25$ as $5^{2}$ .

$x = \frac{- 7 \pm i \cdot \sqrt{5^{2} \cdot 47}}{2 \cdot - 3}$

Step 3.5.1.8

Pull terms out from under the radical.

$x = \frac{- 7 \pm i \cdot (5 \sqrt{47})}{2 \cdot - 3}$

Step 3.5.1.9

Move $5$ to the left of $i$ .

$x = \frac{- 7 \pm 5 i \sqrt{47}}{2 \cdot - 3}$

Step 3.5.2

Multiply $2$ by $- 3$ .

$x = \frac{- 7 \pm 5 i \sqrt{47}}{- 6}$

Step 3.5.3

Simplify $\frac{- 7 \pm 5 i \sqrt{47}}{- 6}$ .

$x = \frac{7 \pm 5 i \sqrt{47}}{6}$

Step 3.6

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 3.6.1

Simplify the numerator.

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Step 3.6.1.1

Raise $7$ to the power of $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot - 3 \cdot - 102}}{2 \cdot - 3}$

Step 3.6.1.2

Multiply $- 4 \cdot - 3 \cdot - 102$ .

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Step 3.6.1.2.1

Multiply $- 4$ by $- 3$ .

$x = \frac{- 7 \pm \sqrt{49 + 12 \cdot - 102}}{2 \cdot - 3}$

Step 3.6.1.2.2

Multiply $12$ by $- 102$ .

$x = \frac{- 7 \pm \sqrt{49 - 1224}}{2 \cdot - 3}$

Step 3.6.1.3

Subtract $1224$ from $49$ .

$x = \frac{- 7 \pm \sqrt{- 1175}}{2 \cdot - 3}$

Step 3.6.1.4

Rewrite $- 1175$ as $- 1 (1175)$ .

$x = \frac{- 7 \pm \sqrt{- 1 \cdot 1175}}{2 \cdot - 3}$

Step 3.6.1.5

Rewrite $\sqrt{- 1 (1175)}$ as $\sqrt{- 1} \cdot \sqrt{1175}$ .

$x = \frac{- 7 \pm \sqrt{- 1} \cdot \sqrt{1175}}{2 \cdot - 3}$

Step 3.6.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 7 \pm i \cdot \sqrt{1175}}{2 \cdot - 3}$

Step 3.6.1.7

Rewrite $1175$ as $5^{2} \cdot 47$ .

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Step 3.6.1.7.1

Factor $25$ out of $1175$ .

$x = \frac{- 7 \pm i \cdot \sqrt{25 (47)}}{2 \cdot - 3}$

Step 3.6.1.7.2

Rewrite $25$ as $5^{2}$ .

$x = \frac{- 7 \pm i \cdot \sqrt{5^{2} \cdot 47}}{2 \cdot - 3}$

Step 3.6.1.8

Pull terms out from under the radical.

$x = \frac{- 7 \pm i \cdot (5 \sqrt{47})}{2 \cdot - 3}$

Step 3.6.1.9

Move $5$ to the left of $i$ .

$x = \frac{- 7 \pm 5 i \sqrt{47}}{2 \cdot - 3}$

Step 3.6.2

Multiply $2$ by $- 3$ .

$x = \frac{- 7 \pm 5 i \sqrt{47}}{- 6}$

Step 3.6.3

Simplify $\frac{- 7 \pm 5 i \sqrt{47}}{- 6}$ .

$x = \frac{7 \pm 5 i \sqrt{47}}{6}$

Step 3.6.4

Change the $\pm$ to $+$ .

$x = \frac{7 + 5 i \sqrt{47}}{6}$

Step 3.7

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 3.7.1

Simplify the numerator.

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Step 3.7.1.1

Raise $7$ to the power of $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot - 3 \cdot - 102}}{2 \cdot - 3}$

Step 3.7.1.2

Multiply $- 4 \cdot - 3 \cdot - 102$ .

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Step 3.7.1.2.1

Multiply $- 4$ by $- 3$ .

$x = \frac{- 7 \pm \sqrt{49 + 12 \cdot - 102}}{2 \cdot - 3}$

Step 3.7.1.2.2

Multiply $12$ by $- 102$ .

$x = \frac{- 7 \pm \sqrt{49 - 1224}}{2 \cdot - 3}$

Step 3.7.1.3

Subtract $1224$ from $49$ .

$x = \frac{- 7 \pm \sqrt{- 1175}}{2 \cdot - 3}$

Step 3.7.1.4

Rewrite $- 1175$ as $- 1 (1175)$ .

$x = \frac{- 7 \pm \sqrt{- 1 \cdot 1175}}{2 \cdot - 3}$

Step 3.7.1.5

Rewrite $\sqrt{- 1 (1175)}$ as $\sqrt{- 1} \cdot \sqrt{1175}$ .

$x = \frac{- 7 \pm \sqrt{- 1} \cdot \sqrt{1175}}{2 \cdot - 3}$

Step 3.7.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 7 \pm i \cdot \sqrt{1175}}{2 \cdot - 3}$

Step 3.7.1.7

Rewrite $1175$ as $5^{2} \cdot 47$ .

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Step 3.7.1.7.1

Factor $25$ out of $1175$ .

$x = \frac{- 7 \pm i \cdot \sqrt{25 (47)}}{2 \cdot - 3}$

Step 3.7.1.7.2

Rewrite $25$ as $5^{2}$ .

$x = \frac{- 7 \pm i \cdot \sqrt{5^{2} \cdot 47}}{2 \cdot - 3}$

Step 3.7.1.8

Pull terms out from under the radical.

$x = \frac{- 7 \pm i \cdot (5 \sqrt{47})}{2 \cdot - 3}$

Step 3.7.1.9

Move $5$ to the left of $i$ .

$x = \frac{- 7 \pm 5 i \sqrt{47}}{2 \cdot - 3}$

Step 3.7.2

Multiply $2$ by $- 3$ .

$x = \frac{- 7 \pm 5 i \sqrt{47}}{- 6}$

Step 3.7.3

Simplify $\frac{- 7 \pm 5 i \sqrt{47}}{- 6}$ .

$x = \frac{7 \pm 5 i \sqrt{47}}{6}$

Step 3.7.4

Change the $\pm$ to $-$ .

$x = \frac{7 - 5 i \sqrt{47}}{6}$

Step 3.8

Identify the leading coefficient.

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Step 3.8.1

The leading term in a polynomial is the term with the highest degree.

$- 3 x^{2}$

Step 3.8.2

The leading coefficient in a polynomial is the coefficient of the leading term.

$- 3$

Step 3.9

Since there are no real x-intercepts and the leading coefficient is negative, the parabola opens down and $- 3 x^{2} + 7 x - 2$ is always less than $0$ .

No solution

Step 4

Find the union of the solutions.

$x < - \frac{14}{3}$ or $x > 7$

Step 5

Convert the inequality to interval notation.

$(- \infty, - \frac{14}{3}) \cup (7, \infty)$

Step 6