# How do you find the zeros, real and imaginary, of y=9x^2-8x-4 using the quadratic formula?

Aug 21, 2017

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:

$\textcolor{red}{9}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 8}$ for $\textcolor{b l u e}{b}$

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

$x = \frac{- \textcolor{b l u e}{\left(- 8\right)} \pm \sqrt{{\textcolor{b l u e}{\left(- 8\right)}}^{2} - \left(4 \cdot \textcolor{red}{9} \cdot \textcolor{g r e e n}{- 4}\right)}}{2 \cdot \textcolor{red}{9}}$

$x = \frac{8 \pm \sqrt{64 - \left(- 144\right)}}{18}$

$x = \frac{8 \pm \sqrt{64 + \left(144\right)}}{18}$

$x = \frac{8 \pm \sqrt{208}}{18}$

$x = \frac{8 \pm \sqrt{16 \cdot 13}}{18}$

$x = \frac{8 \pm \sqrt{16} \sqrt{13}}{18}$

$x = \frac{8 \pm 4 \sqrt{13}}{18}$

$x = \frac{8 - 4 \sqrt{13}}{18}$ and $x = \frac{8 + 4 \sqrt{13}}{18}$

$x = \frac{4 - 2 \sqrt{13}}{9}$ and $x = \frac{4 + 2 \sqrt{13}}{9}$