# How do you solve x^2 + 4x - 12= 0 using completing the square?

Jun 11, 2015

$0 = {x}^{2} + 4 x - 12 = {\left(x + 2\right)}^{2} - 4 - 12 = {\left(x + 2\right)}^{2} - 16$
$= {\left(x + 2\right)}^{2} - {4}^{2}$

Hence $x = - 2 \pm 4$

That is $x = - 6$ or $x = 2$

#### Explanation:

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

So:

$0 = {x}^{2} + 4 x - 12 = {x}^{2} + 4 x + 4 - 4 - 12$

$= {\left(x + 2\right)}^{2} - 4 - 12 = {\left(x + 2\right)}^{2} - 16 = {\left(x + 2\right)}^{2} - {4}^{2}$

Add ${4}^{2}$ to both ends to get:

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

Hence:

$x + 2 = \pm 4$

Subtract $2$ from both sides to get:

$x = - 2 \pm 4$