# How do you solve 2/3 x^2 + 1/4 x= 3?

##### 1 Answer
May 17, 2018

See a solution process below:

#### Explanation:

First, multiply each side of the equation by $\textcolor{red}{12}$ to eliminate the fractions:

$\textcolor{red}{12} \left(\frac{2}{3} {x}^{2} + \frac{1}{4} x\right) = \textcolor{red}{12} \times 3$

$\left(\textcolor{red}{12} \times \frac{2}{3} {x}^{2}\right) + \left(\textcolor{red}{12} \times \frac{1}{4} x\right) = 36$

$8 {x}^{2} + 3 x = 36$

Next, put the equation in standard quadratic form:

$8 {x}^{2} + 3 x - \textcolor{red}{36} = 36 - \textcolor{red}{36}$

$8 {x}^{2} + 3 x - 36 = 0$

We can nowuse the quadratic equation to solve this problem:

The quadratic formula states:

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:

$\textcolor{red}{8}$ for $\textcolor{red}{a}$

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

$\textcolor{g r e e n}{- 36}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{3} \pm \sqrt{{\textcolor{b l u e}{3}}^{2} - \left(4 \cdot \textcolor{red}{8} \cdot \textcolor{g r e e n}{- 36}\right)}}{2 \cdot \textcolor{red}{8}}$

$x = \frac{- \textcolor{b l u e}{3} \pm \sqrt{9 - \left(- 1152\right)}}{16}$

$x = \frac{- \textcolor{b l u e}{3} \pm \sqrt{9 + 1152}}{16}$

$x = \frac{- \textcolor{b l u e}{3} \pm \sqrt{1161}}{16}$

$x = \frac{- \textcolor{b l u e}{3} \pm \sqrt{9 \cdot 129}}{16}$

$x = \frac{- \textcolor{b l u e}{3} \pm \sqrt{9} \sqrt{129}}{16}$

$x = \frac{- \textcolor{b l u e}{3} \pm 3 \sqrt{129}}{16}$

$x = \frac{3 \left(- 1 \pm \sqrt{129}\right)}{16}$

Or

$x = \frac{3}{16} \left(- 1 \pm \sqrt{129}\right)$