# How do you solve -15x^2 - 10x + 40 = 0  using the quadratic formula?

Aug 14, 2015

${x}_{1 , 2} = - \frac{1 \pm 5}{3}$

#### Explanation:

For a general form quadratic equation

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

the two possible roots of the equation take the form - this is the quadratic formula

color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a)

Now, your quadratic equation looks like this

$- 15 {x}^{2} - 10 x + 40 = 0$

You can divide all the terms by $5$ to simplify it

$- 3 {x}^{2} - 2 x + 8 = 0$

In your case, you have $a = - 3$, $b = - 2$, and $c = 8$. THis means that the two solutions will be

${x}_{1 , 2} = \frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \cdot \left(- 3\right) \cdot 8}}{2 \cdot \left(- 3\right)}$

${x}_{1 , 2} = \frac{2 \pm \sqrt{100}}{- 6}$

${x}_{1 , 2} = \frac{2 \pm 10}{- 6} = - \frac{1 \pm 5}{3} = \left\{\begin{matrix}{x}_{1} = \frac{- 1 - 5}{3} = - 2 \\ {x}_{2} = \frac{- 1 + 5}{3} = \frac{4}{3}\end{matrix}\right.$

Your quadratic equation will thus have two roots, ${x}_{1} = \textcolor{g r e e n}{- 2}$ and ${x}_{2} = \textcolor{g r e e n}{\frac{4}{3}}$.