# How do you solve 0=2x^2 + 5x - 12 using the quadratic formula?

Oct 19, 2015

${x}_{1 , 2} = \frac{- 5 \pm 11}{4}$

#### Explanation:

For a general form quadratic equation

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

its roots can be determined using the quadratic formula

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

$2 {x}^{2} + 5 x - 12 = 0$

which implies that $a = 2$, $b = 5$, and $c = - 12$.

The two roots will thus take the form

${x}_{1 , 2} = \frac{- 5 \pm \sqrt{{5}^{2} - 4 \cdot 2 \cdot \left(- 12\right)}}{2 \cdot 2}$

${x}_{1 , 2} = \frac{- 5 \pm \sqrt{121}}{4}$

${x}_{1 , 2} = \frac{- 5 \pm 11}{4} = \left\{\begin{matrix}{x}_{1.} = \frac{_ 5 - 11}{4} = - 4 \\ {x}_{2} = \frac{- 5 + 11}{4} = \frac{3}{2}\end{matrix}\right.$

The two solutions to this quadratic equation will thus be

${x}_{1} = \textcolor{g r e e n}{- 16} \text{ }$ and $\text{ } {x}_{2} = \textcolor{g r e e n}{\frac{3}{2}}$