# How do you find the zeros, real and imaginary, of y=-2x^2-19x+42 using the quadratic formula?

Oct 21, 2017

See a solution process below:

#### Explanation:

To find the zeros of a quadratic equation you set the quadratic equal to $0$:

$0 = - 2 {x}^{2} - 19 x + 42$

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:

$\textcolor{red}{- 2}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 19}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{42}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{\left(- 19\right)} \pm \sqrt{{\textcolor{b l u e}{\left(- 19\right)}}^{2} - \left(4 \cdot \textcolor{red}{- 2} \cdot \textcolor{g r e e n}{42}\right)}}{2 \cdot \textcolor{red}{- 2}}$

$x = \frac{\textcolor{b l u e}{19} \pm \sqrt{361 - \left(- 336\right)}}{-} 4$

$x = \frac{\textcolor{b l u e}{19} \pm \sqrt{361 + 336}}{-} 4$

$x = \frac{\textcolor{b l u e}{19} \pm \sqrt{697}}{-} 4$

$x = - \frac{\textcolor{b l u e}{19} \pm \sqrt{697}}{4}$