# How do you solve 2x^2-3x-6=0 using the quadratic formula?

Aug 28, 2017

See a solution process below:

#### Explanation:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{2}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 3}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 6}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{\left(- 3\right)} \pm \sqrt{{\textcolor{b l u e}{\left(- 3\right)}}^{2} - \left(4 \cdot \textcolor{red}{2} \cdot \textcolor{g r e e n}{- 6}\right)}}{2 \cdot \textcolor{red}{2}}$

$x = \frac{\textcolor{b l u e}{3} \pm \sqrt{9 - \left(- 48\right)}}{4}$

$x = \frac{\textcolor{b l u e}{3} \pm \sqrt{9 + 48}}{4}$

$x = \frac{\textcolor{b l u e}{3} \pm \sqrt{57}}{4}$