Finite Math Examples

$x^{2} + 4 = - 3 x$

Step 1

Add $3 x$ to both sides of the equation.

$x^{2} + 4 + 3 x = 0$

Step 2

Reorder terms.

$x^{2} + 3 x + 4 = 0$

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

Substitute the values $a = 1$ , $b = 3$ , and $c = 4$ into the quadratic formula and solve for $x$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (1 \cdot 4)}}{2 \cdot 1}$

Step 5

Simplify.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

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

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

Step 5.1.2

Multiply $- 4 \cdot 1 \cdot 4$ .

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Step 5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 4}}{2 \cdot 1}$

Step 5.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{- 3 \pm \sqrt{9 - 16}}{2 \cdot 1}$

Step 5.1.3

Subtract $16$ from $9$ .

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

Step 5.1.4

Rewrite $- 7$ as $- 1 (7)$ .

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

Step 5.1.5

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

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

Step 5.1.6

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

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

Step 5.2

Multiply $2$ by $1$ .

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

Step 6

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

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Step 6.1

Simplify the numerator.

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Step 6.1.1

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

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

Step 6.1.2

Multiply $- 4 \cdot 1 \cdot 4$ .

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Step 6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 4}}{2 \cdot 1}$

Step 6.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{- 3 \pm \sqrt{9 - 16}}{2 \cdot 1}$

Step 6.1.3

Subtract $16$ from $9$ .

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

Step 6.1.4

Rewrite $- 7$ as $- 1 (7)$ .

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

Step 6.1.5

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

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

Step 6.1.6

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

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

Step 6.2

Multiply $2$ by $1$ .

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

Step 6.3

Change the $\pm$ to $+$ .

$x = \frac{- 3 + i \sqrt{7}}{2}$

Step 6.4

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \cdot 3 + i \sqrt{7}}{2}$

Step 6.5

Factor $- 1$ out of $i \sqrt{7}$ .

$x = \frac{- 1 \cdot 3 - (- i \sqrt{7})}{2}$

Step 6.6

Factor $- 1$ out of $- 1 (3) - (- i \sqrt{7})$ .

$x = \frac{- 1 (3 - i \sqrt{7})}{2}$

Step 6.7

Move the negative in front of the fraction.

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

Step 7

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

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Step 7.1

Simplify the numerator.

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Step 7.1.1

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

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

Step 7.1.2

Multiply $- 4 \cdot 1 \cdot 4$ .

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Step 7.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 4}}{2 \cdot 1}$

Step 7.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{- 3 \pm \sqrt{9 - 16}}{2 \cdot 1}$

Step 7.1.3

Subtract $16$ from $9$ .

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

Step 7.1.4

Rewrite $- 7$ as $- 1 (7)$ .

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

Step 7.1.5

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

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

Step 7.1.6

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

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

Step 7.2

Multiply $2$ by $1$ .

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

Step 7.3

Change the $\pm$ to $-$ .

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

Step 7.4

Rewrite $- 3$ as $- 1 (3)$ .

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

Step 7.5

Factor $- 1$ out of $- i \sqrt{7}$ .

$x = \frac{- 1 \cdot 3 - (i \sqrt{7})}{2}$

Step 7.6

Factor $- 1$ out of $- 1 (3) - (i \sqrt{7})$ .

$x = \frac{- 1 (3 + i \sqrt{7})}{2}$

Step 7.7

Move the negative in front of the fraction.

$x = - \frac{3 + i \sqrt{7}}{2}$

Step 8

The final answer is the combination of both solutions.

$x = - \frac{3 - i \sqrt{7}}{2}, - \frac{3 + i \sqrt{7}}{2}$