# How do you solve 3x^2 + 13x = 10?

Mar 26, 2018

See a solution process below:

#### Explanation:

First, subtract $\textcolor{red}{10}$ from each side of the equation to put the equation in standard quadratic for while keeping the equation balanced:

$3 {x}^{2} + 13 x - \textcolor{red}{10} = 10 - \textcolor{red}{10}$

$3 {x}^{2} + 13 x - 10 = 0$

We can now 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:

$\textcolor{red}{3}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{13}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 10}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{13} \pm \sqrt{{\textcolor{b l u e}{13}}^{2} - \left(4 \cdot \textcolor{red}{3} \cdot \textcolor{g r e e n}{- 10}\right)}}{2 \cdot \textcolor{red}{3}}$

$x = \frac{- \textcolor{b l u e}{13} \pm \sqrt{169 - \left(12 \cdot \textcolor{g r e e n}{- 10}\right)}}{6}$

$x = \frac{- \textcolor{b l u e}{13} \pm \sqrt{169 - \left(- 120\right)}}{6}$

$x = \frac{- \textcolor{b l u e}{13} \pm \sqrt{169 + 120}}{6}$

$x = \frac{- \textcolor{b l u e}{13} \pm \sqrt{289}}{6}$

$x = \frac{- \textcolor{b l u e}{13} - 17}{6}$ and $x = \frac{- \textcolor{b l u e}{13} + 17}{6}$

$x = - \frac{30}{6}$ and $x = \frac{4}{6}$

$x = - 5$ and $x = \frac{2}{3}$

The Solution Set Is: $x = \left\{- 5 , \frac{2}{3}\right\}$