# How do you solve x^2+6x+9=-32?

Apr 23, 2016

$x = - 3 \pm 4 \sqrt{2} i$

#### Explanation:

The left hand side here is a perfect square trinomial, but the right hand side is negative, so only has an imaginary square root.

We find:

${x}^{2} + 6 x + 9 = {\left(x + 3\right)}^{2}$

Hence:

${\left(x + 3\right)}^{2} = - 32$

So:

$x + 3 = \pm \sqrt{- 32} = \pm \sqrt{32} i = \pm 4 \sqrt{2} i$

Subtract $3$ from both sides to get:

$x = - 3 \pm 4 \sqrt{2} i$