Algebra Examples

$2 x^{2} + 12 x = - 10$

Step 1

Divide each term in $2 x^{2} + 12 x = - 10$ by $2$ and simplify.

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Step 1.1

Divide each term in $2 x^{2} + 12 x = - 10$ by $2$ .

$\frac{2 x^{2}}{2} + \frac{12 x}{2} = \frac{- 10}{2}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Simplify each term.

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Step 1.2.1.1

Cancel the common factor of $2$ .

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Step 1.2.1.1.1

Cancel the common factor.

$\frac{2 x^{2}}{2} + \frac{12 x}{2} = \frac{- 10}{2}$

Step 1.2.1.1.2

Divide $x^{2}$ by $1$ .

$x^{2} + \frac{12 x}{2} = \frac{- 10}{2}$

Step 1.2.1.2

Cancel the common factor of $12$ and $2$ .

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Step 1.2.1.2.1

Factor $2$ out of $12 x$ .

$x^{2} + \frac{2 (6 x)}{2} = \frac{- 10}{2}$

Step 1.2.1.2.2

Cancel the common factors.

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Step 1.2.1.2.2.1

Factor $2$ out of $2$ .

$x^{2} + \frac{2 (6 x)}{2 (1)} = \frac{- 10}{2}$

Step 1.2.1.2.2.2

Cancel the common factor.

$x^{2} + \frac{2 (6 x)}{2 \cdot 1} = \frac{- 10}{2}$

Step 1.2.1.2.2.3

Rewrite the expression.

$x^{2} + \frac{6 x}{1} = \frac{- 10}{2}$

Step 1.2.1.2.2.4

Divide $6 x$ by $1$ .

$x^{2} + 6 x = \frac{- 10}{2}$

Step 1.3

Simplify the right side.

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Step 1.3.1

Divide $- 10$ by $2$ .

$x^{2} + 6 x = - 5$

Step 2

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(3)}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} + 6 x + {(3)}^{2} = - 5 + {(3)}^{2}$

Step 4

Simplify the equation.

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Step 4.1

Simplify the left side.

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Step 4.1.1

Raise $3$ to the power of $2$ .

$x^{2} + 6 x + 9 = - 5 + {(3)}^{2}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $- 5 + {(3)}^{2}$ .

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Step 4.2.1.1

Raise $3$ to the power of $2$ .

$x^{2} + 6 x + 9 = - 5 + 9$

Step 4.2.1.2

Add $- 5$ and $9$ .

$x^{2} + 6 x + 9 = 4$

Step 5

Factor the perfect trinomial square into ${(x + 3)}^{2}$ .

${(x + 3)}^{2} = 4$

Step 6

Solve the equation for $x$ .

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Step 6.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x + 3 = \pm \sqrt{4}$

Step 6.2

Simplify $\pm \sqrt{4}$ .

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Step 6.2.1

Rewrite $4$ as $2^{2}$ .

$x + 3 = \pm \sqrt{2^{2}}$

Step 6.2.2

Pull terms out from under the radical, assuming positive real numbers.

$x + 3 = \pm 2$

Step 6.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 6.3.1

First, use the positive value of the $\pm$ to find the first solution.

$x + 3 = 2$

Step 6.3.2

Move all terms not containing $x$ to the right side of the equation.

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Step 6.3.2.1

Subtract $3$ from both sides of the equation.

$x = 2 - 3$

Step 6.3.2.2

Subtract $3$ from $2$ .

$x = - 1$

Step 6.3.3

Next, use the negative value of the $\pm$ to find the second solution.

$x + 3 = - 2$

Step 6.3.4

Move all terms not containing $x$ to the right side of the equation.

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Step 6.3.4.1

Subtract $3$ from both sides of the equation.

$x = - 2 - 3$

Step 6.3.4.2

Subtract $3$ from $- 2$ .

$x = - 5$

Step 6.3.5

The complete solution is the result of both the positive and negative portions of the solution.

$x = - 1, - 5$