# How do you factor -x^{2}-12x-18=0?

Apr 22, 2018

The solution is x=-6±3sqrt2

and hence the factorization is
(x+6-3sqrt2)(x+6+3sqrt2))=0

#### Explanation:

$- {x}^{2} - 12 x - 18 = 0$
${x}^{2} + 12 x + 18 = 0$

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

$x = \frac{- 12 \pm \sqrt{144 - 72}}{2 \times 1}$

$x = \frac{- 12 \pm \sqrt{72}}{2}$

$x = \frac{- 12 \pm 6 \sqrt{2}}{2}$

x=-6±3sqrt2

These are the solutions when $- {x}^{2} - 12 x - 18 = 0$

The factorization can be written in the form $\left(x - a\right) \left(x - b\right)$ where $a$ and $b$ are the solutions, also known as zeros.

$\left(x - \left(- 6 + 3 \sqrt{2}\right)\right) \left(x - \left(- 6 - 3 \sqrt{2}\right)\right) = 0$

(x+6-3sqrt2)(x+6+3sqrt2))=0