# How do you solve x^2-3x+9/4=0 using the quadratic formula?

May 28, 2018

See a solution process below:

#### Explanation:

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:

$\textcolor{red}{1}$ for $\textcolor{red}{a}$

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

$\textcolor{g r e e n}{\frac{9}{4}}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{- 3} \pm \sqrt{{\textcolor{b l u e}{- 3}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{\frac{9}{4}}\right)}}{2 \cdot \textcolor{red}{1}}$

$x = \frac{3 \pm \sqrt{9 - 9}}{2}$

$x = \frac{3 \pm 0}{2}$

$x = \frac{3}{2}$