# How do you solve the quadratic equation by completing the square: x^2+2x-5=0?

Aug 1, 2015

${x}_{1 , 2} = - 1 \pm \sqrt{6}$

#### Explanation:

$\textcolor{b l u e}{{x}^{2} + \frac{b}{a} x = - \frac{c}{a}}$

To do that, add $5$ to both sides of the equation

${x}^{2} + 2 x - \textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} = 5$

${x}^{2} + 2 x = 5$

Next, divide the coefficient of the $x$-term by 2, square the result. then add it to both sides of the equation.

$\frac{2}{2} = 1$, then ${1}^{2} = 1$

This will get you

${x}^{2} + 2 x + 1 = 5 + 1$

Notice that you can rewrite the left side of the equation as the square of a binomial

${x}^{2} + 2 x + 1 = {x}^{2} + 2 \cdot \left(1\right) \cdot x + {1}^{2} = {\left(x + 1\right)}^{2}$

You will now have

${\left(x + 1\right)}^{2} = 6$

Take the square root of both sides

$\sqrt{{\left(x + 1\right)}^{2}} = \sqrt{6}$

$x + 1 = \pm \sqrt{6} \implies {x}_{1 , 2} = - 1 \pm \sqrt{6} = \left\{\begin{matrix}{x}_{1} = \textcolor{g r e e n}{- 1 - \sqrt{6}} \\ {x}_{2} = \textcolor{g r e e n}{- 1 + \sqrt{6}}\end{matrix}\right.$