# How do you solve the quadratic equation by completing the square: 5x^2 + 8x - 2 = 0?

Aug 5, 2015

${x}_{1 , 2} = - \frac{4}{5} \pm \frac{\sqrt{26}}{5}$

#### Explanation:

To solve this quadratic by completing the square, you first need to get your quadratic to the form

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

In your case, this means adding $2$ to both sides of the equation and dividing everything by $5$

$5 {x}^{2} + 8 x - \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} = 0 + 2$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} {x}^{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}}} + \frac{8}{5} x = \frac{2}{5}$

${x}^{2} + \frac{8}{5} x = \frac{2}{5}$

Next, use the coefficient of the $x$-term to help you determine what term needs to be added to both sides of the equation so that the left side can be written as the square of a binomial.

More specifically, divide said coefficient by $2$, square the result, then add it to both sides of the equation

${\left(\frac{8}{5} \cdot \frac{1}{2}\right)}^{2} = \frac{16}{25}$

Your equation will now look like this

${x}^{2} + \frac{8}{5} x + \frac{16}{25} = \frac{2}{5} + \frac{16}{25}$

The left side of the equation can be written as

${x}^{2} + 2 \cdot \left(\frac{4}{5}\right) \cdot x + {\left(\frac{4}{5}\right)}^{2} = {\left(x + \frac{4}{5}\right)}^{2}$

This means that you now have

${\left(x + \frac{4}{5}\right)}^{2} = \frac{26}{25}$

Take the square root of both sides to get

$\sqrt{{\left(x + \frac{4}{5}\right)}^{2}} = \sqrt{\frac{26}{25}}$

$x + \frac{4}{5} = \pm \frac{\sqrt{26}}{5} \implies {x}_{1 , 2} = - \frac{4}{5} \pm \frac{\sqrt{26}}{5}$

${x}_{1} = \textcolor{g r e e n}{- \frac{4}{5} - \frac{\sqrt{26}}{5}}$ and ${x}_{2} = \textcolor{g r e e n}{- \frac{4}{5} + \frac{\sqrt{26}}{5}}$