# How do you solve 4x^2=9x using the quadratic formula?

Aug 21, 2017

See a solution process below:

#### Explanation:

First, rewrite the equation as:

$4 {x}^{2} - 9 x + 0 = 0$

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:

$\textcolor{red}{4}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 9}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{0}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{\left(- 9\right)} \pm \sqrt{{\textcolor{b l u e}{\left(- 9\right)}}^{2} - \left(4 \cdot \textcolor{red}{4} \cdot \textcolor{g r e e n}{0}\right)}}{2 \cdot \textcolor{red}{4}}$

$x = \frac{9 \pm \sqrt{81 - 0}}{8}$

$x = \frac{9 \pm \sqrt{81}}{8}$

$x = \frac{9 - 9}{8}$ and $x = \frac{9 + 9}{8}$

$x = \frac{0}{8}$ and $x = \frac{18}{8}$

$x = 0$ and $x = \frac{9}{4}$