# How do you solve x^2+4x-3=0 by completing the square?

May 11, 2015

$0 = {x}^{2} + 4 x - 3 = {x}^{2} + 4 x + 4 - 7 = {\left(x + 2\right)}^{2} - 7$

Adding 7 to both ends we get:

${\left(x + 2\right)}^{2} = 7$

So $x + 2 = \pm \sqrt{7}$

Subtracting 2 from both sides, we arrive at:

$x = \pm \sqrt{7} - 2$.

In general, $a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$.