# How do you solve 2p^2 - 3p -2 = 0?

Mar 7, 2018

See a solution process below:

#### Explanation:

We can use the quadratic equation to solve this problem:

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 $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 2}$ for $\textcolor{g r e e n}{c}$ gives:

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

$p = \frac{\textcolor{b l u e}{3} \pm \sqrt{9 - \left(- 16\right)}}{4}$

$p = \frac{\textcolor{b l u e}{3} \pm \sqrt{9 + 16}}{4}$

$p = \frac{\textcolor{b l u e}{3} \pm \sqrt{25}}{4}$

$p = \frac{\textcolor{b l u e}{3} - 5}{4}$; $p = \frac{\textcolor{b l u e}{3} + 5}{4}$

$p = - \frac{2}{4}$; $p = \frac{8}{4}$

$p = - \frac{1}{2}$; $p = 2$