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

Jul 5, 2015

Force a perfect square trinomial on the left side of the equation. Take the square root of both sides and solve for $x$.

#### Explanation:

${x}^{2} + 6 x - 4 = 0$

Force a perfect square trinomial such that ${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$ on the left side of the equation. Then take the square root of both sides and solve for $x$.

Add $4$ to both sides of the equation.

${x}^{2} + 6 x = 4$

Take the coefficient of the $x$ term and divide it by $2$ and square the result.

6/2=3; ${3}^{2} = 9$

Add $9$ to both sides.

${x}^{2} + 6 x + 9 = 4 + 9$ =

${x}^{2} + 6 x + 9 = 13$

a=x; $b = 3$

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

Take the square root of both sides.

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

Subtract $3$ from both sides.

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

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