# How do you solve the quadratic equation by completing the square: x^2+6x-4=0?

Jul 21, 2015

Add $13$ to both sides to make the left hand side a perfect square trinomial, then take the square root of both sides and subtract $3$ to find:

$x = - 3 \pm \sqrt{13}$

#### Explanation:

Add $13$ to both sides to get:

$13 = {x}^{2} + 6 x + 9 = {\left(x + 3\right)}^{2}$

${\left(x + 3\right)}^{2} = 13$

Take the square root of both sides to get rid of the exponent:

$\rightarrow \sqrt{{\left(x + 3\right)}^{2}} = \pm \sqrt{13}$

So $x + 3 = \pm \sqrt{13}$

Subtract $3$ from both sides to get:

$x = - 3 \pm \sqrt{13}$