# How do you find all the real and complex roots of z² + 4z + 8 = 0?

Feb 6, 2016

$z = - 2 \pm 2 i$

#### Explanation:

Complete the square:

${z}^{2} + 4 z = - 8$

Add $4$ to both sides since ${x}^{2} + 4 z + 4 = {\left(z + 2\right)}^{2}$.

${z}^{2} + 4 z + 4 = - 8 + 4$

${\left(z + 2\right)}^{2} = - 4$

Take the square root of both sides.

$z + 2 = \pm \sqrt{- 4}$

$z + 2 = \pm 2 i$

$z = - 2 \pm 2 i$

This can also be solved through the quadratic formula:

$z = \frac{- 4 \pm \sqrt{16 - 32}}{2} = \frac{- 4 \pm 4 i}{2} = - 2 \pm 2 i$