# How do you find the zeros, if any, of y= -8x^2 -x-2using the quadratic formula?

May 24, 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:

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

$\textcolor{b l u e}{- 1}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 2}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{- 1} \pm \sqrt{{\textcolor{b l u e}{- 1}}^{2} - \left(4 \cdot \textcolor{red}{- 8} \cdot \textcolor{g r e e n}{- 2}\right)}}{2 \cdot \textcolor{red}{- 8}}$

$x = \frac{1 \pm \sqrt{1 - \left(64\right)}}{-} 16$

$x = \frac{1 \pm \sqrt{- 63}}{-} 16$

Because the square root of a negative number is not a Real Number there are no zeros for this equation.

Graphing this equation shows:

graph{(y+8x^2+x+2)=0 [-20, 20, -15, 5]}