# How do you solve x^2 + 6x + 7 = 0 by completing the square?

May 5, 2015

Note that $\textcolor{red}{{x}^{2} + 6 x + 9} = \textcolor{b l u e}{{\left(x + 3\right)}^{2}}$

Given ${x}^{2} + 6 x + 7 = 0$
we can rewrite this as
$\textcolor{red}{{x}^{2} + 6 x + 9} - 2 = 0$

$\textcolor{b l u e}{{\left(x + 3\right)}^{2}} = 2$

$x + 3 = \pm \sqrt{2}$

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