# How do you solve the quadratic equation by completing the square: x^2+4x=5?

Jul 19, 2015

Complete the square by adding $4$ to both sides of the equation to find:

${\left(x + 2\right)}^{2} = {x}^{2} + 4 x + 4 = 5 + 4 = 9 = {3}^{2}$

Hence $x = 1$ or $x = - 5$

#### Explanation:

Given $a {x}^{2} + b x$, note that

$a {\left(x + \frac{b}{2 a}\right)}^{2} = a {x}^{2} + b x + {b}^{2} / \left(4 a\right)$

In our case $a = 1$ and $b = 4$, so $\frac{b}{2 a} = \frac{4}{2} = 2$ and

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

So add $4$ to both sides of our original equation to get:

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

That is:

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

So

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

Subtract $2$ from both sides to get:

$x = - 2 \pm 3$