# How do you complete the square to solve x^2+2x-10=14?

May 23, 2015

If you first add $11$ to both sides you get:

${x}^{2} + 2 x + 1 = 25$

Now

${x}^{2} + 2 x + 1 = \left(x + 1\right) \left(x + 1\right) = {\left(x + 1\right)}^{2}$

and $25 = {5}^{2}$

So we have

${\left(x + 1\right)}^{2} = {5}^{2}$

Hence

$x + 1 = \pm \sqrt{{5}^{2}} = \pm 5$

Subtracting 1 from both sides we get:

$x = - 1 \pm 5$

That is $x = 4$ or $x = - 6$