# How do you solve x^2 + 8x + 13 = 0?

Feb 25, 2016

$x = - 4 + \sqrt{3} , - 4 - \sqrt{3}$

#### Explanation:

${x}^{2} + 8 x + 13 = 0$ is a quadratic equation in standard form $a {x}^{2} + b x + 13$, where $a = 1 , b = 8 , c = 13$.

The quadratic formula can be used to solve the equation.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Substitute the known values into the formula and solve for $x$.

$x = \frac{- 8 \pm \sqrt{{8}^{2} - \left(4 \cdot 1 \cdot 13\right)}}{2 \cdot 1}$

Simplify.

$x = \frac{- 8 \pm \sqrt{64 - 52}}{2}$

Simplify.

$x = \frac{- 8 \pm \sqrt{12}}{2}$

Factor $\sqrt{12}$.

$\sqrt{2 \times 2 \times 3} =$

$\sqrt{{2}^{2} \times 3} =$

$2 \sqrt{3}$

$x = \frac{- 8 \pm 2 \sqrt{3}}{2}$

Simplify.

$x = \frac{{\cancel{- 8}}^{-} 4 \pm {\cancel{2}}^{1} \sqrt{3}}{\cancel{2}} ^ 1$

Simplify.

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

Solutions for $x$.

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

$x = - 4 - \sqrt{3}$