# How do you solve 3x^2 + 11x – 20 = 0 using the quadratic formula?

Jul 27, 2018

$x = \frac{4}{3}$ and $x = - 5$

#### Explanation:

We know that the quadratic formula is $\frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{g r e e n}{c}}}{2 \textcolor{red}{a}}$.
All we need to do is extract the values for $\textcolor{red}{a} , \textcolor{b l u e}{b} ,$ and $\textcolor{g r e e n}{c}$ from the equation.

$\textcolor{red}{3} {x}^{2} + \textcolor{b l u e}{11} x \textcolor{g r e e n}{- 20}$

We can see that:
color(red)(a)=color(red)(3
color(blue)(b)=color(blue)(11
color(green)(c)=color(green)(-20

Now, we can just plug those values into the quadratic formula:

$\frac{- \textcolor{b l u e}{11} \pm \sqrt{{\textcolor{b l u e}{11}}^{2} - 4 \textcolor{red}{\left(3\right)} \textcolor{g r e e n}{\left(- 20\right)}}}{2 \textcolor{red}{\left(3\right)}}$

$\frac{- 11 \pm \sqrt{121 + 240}}{6}$

$\frac{- 11 \pm \sqrt{361}}{6}$

$\frac{- 11 \pm 19}{6}$, which is

$\frac{- 11 + 19}{6}$ and $\frac{- 11 - 19}{6}$, which is

$\frac{8}{6}$ and $- \frac{30}{6}$, which is

$\frac{4}{3}$ and $- 5$.

These are the solutions to the equation above.