# How do you solve w^2+4w-2=0 by completing the square?

May 23, 2015

$0 = {w}^{2} + 4 w - 2 = {w}^{2} + 4 w + 4 - 4 - 2 = {\left(w + 2\right)}^{2} - 6$

Add $6$ to both ends to get:

${\left(w + 2\right)}^{2} = 6$

So

$w + 2 = \pm \sqrt{6}$

Subtract $2$ from both sides to get:

$w = - 2 \pm \sqrt{6}$