# How do you solve 2x^2 + 6x = 5 by using the quadratic formula?

Oct 20, 2015

${x}_{1 , 2} = \frac{- 6 \pm \sqrt{76}}{4}$

#### Explanation:

For a general form quadratic equation

$\textcolor{b l u e}{a {x}^{2} + b x + c = 0}$

you can calculate the roots by using the quadratic formula

$\textcolor{b l u e}{{x}_{1 , 2} = \frac{_ b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}}$

So, the first thing to do here is rearrange your equation to match the general form. To do that, add $- 5$ to both sides of the equation

$2 {x}^{2} + 6 x - 5 = \textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{5}}}$

$2 {x}^{2} + 6 x - 5 = 0$

In your case, you have $a = 2$, $b = 6$, and $c = - 5$. This means that the two solutions will take the form

${x}_{1 , 2} = \frac{- 6 \pm \sqrt{{6}^{2} - 4 \cdot 2 \cdot \left(- 5\right)}}{2 \cdot 2}$

${x}_{1 , 2} = \frac{- 6 \pm \sqrt{76}}{4}$

SInce $76$ is not a perfect square, the two solutions will be

${x}_{1} = \frac{- 6 - \sqrt{76}}{4} \text{ }$ and $\text{ } {x}_{2} = \frac{- 6 \pm \sqrt{76}}{4}$

You an try out one to see if the calculations are correct

$2 \cdot {\left(\frac{- 6 - \sqrt{76}}{4}\right)}^{2} + 6 \cdot \left(\frac{- 6 - \sqrt{76}}{4}\right) - 5 = 0$

$\frac{2}{16} \cdot \left(36 + 2 \sqrt{76} + 76\right) + \frac{3}{2} \cdot \left(- 6 - \sqrt{76}\right) - 5 = 0$

$14 + \textcolor{red}{\cancel{\textcolor{b l a c k}{\frac{3}{2} \sqrt{76}}}} - 9 - \textcolor{red}{\cancel{\textcolor{b l a c k}{\frac{3}{2} \sqrt{76}}}} - 5 = 0$

$14 - 9 - 5 = 0 \textcolor{w h i t e}{x} \textcolor{g r e e n}{\sqrt{}}$