# How do you complete the square to solve x^2-2x-3=0?

May 24, 2015

$0 = {x}^{2} - 2 x - 3 = {x}^{2} - 2 x + 1 - 1 - 3 = {\left(x - 1\right)}^{2} - 4$

Adding $4$ to both ends we find:

${\left(x - 1\right)}^{2} = 4$

So $x - 1 = \pm \sqrt{4} = \pm 2$

Adding $1$ to both sides, we get:

$x = 1 \pm 2$

That is $x = - 1$ or $x = 3$