# How do you solve -10x^2 + 11x + 24 = 20 using the quadratic formula?

Aug 30, 2015

${x}_{1 , 2} = \frac{11 \pm \sqrt{281}}{20}$

#### Explanation:

For a general form quadratic equation

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

you can use the quadratic formula to determine the roots of the equation

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

So, start by getting your equation into standard quadratic form. To do that, add $- 20$ to both sides of the equation

$- 10 {x}^{2} + 11 x + 24 - 20 = \textcolor{red}{\cancel{\textcolor{b l a c k}{20}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{20}}}$

$- 10 {x}^{2} + 11 x + 4 = 0$

In your case, you have $a = - 10$, $b = 11$, and $c = 4$, which means that the quadratic formula will look like this

${x}_{1 , 2} = \frac{- 11 \pm \sqrt{{11}^{2} - 4 \cdot \left(- 10\right) \cdot \left(4\right)}}{2 \cdot \left(- 10\right)}$

${x}_{1 , 2} = \frac{- 11 \pm \sqrt{281}}{\left(- 20\right)} = \frac{11 \pm \sqrt{281}}{20}$

The two roots of the quadratic equation will thus be

${x}_{1} = \frac{11 + \sqrt{281}}{20} \text{ }$ and $\text{ } {x}_{2} = \frac{11 - \sqrt{281}}{20}$