# How do you solve x^2+6x=7  using completing the square?

Jun 19, 2015

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

So: ${\left(x + 3\right)}^{2} = 7 + 9 = 16 = {4}^{2}$

Hence $x = - 7$ or $x = 1$

#### Explanation:

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

So: ${\left(x + 3\right)}^{2} = {x}^{2} + 6 x + 9 = 7 + 9 = 16 = {4}^{2}$

Take square root of both ends, allowing for both possibilities:

$x + 3 = \pm \sqrt{{4}^{2}} = \pm 4$

Subtract $3$ from both sides to get:

$x = - 3 \pm 4$

That is $x = - 7$ or $x = 1$