# How do you find the complex roots of b^2+36=0?

Dec 16, 2016

$b = \pm 6 i$

#### Explanation:

rearrange as normal

${b}^{2} = - 36$

$b = \sqrt{- 36}$

$b = \pm \sqrt{- 1} \sqrt{36}$

$b = \pm 6 i$