# How do you find the zeros, real and imaginary, of y=-x^2-3x+11 using the quadratic formula?

Oct 18, 2017

See a solution process below:

#### Explanation:

We can use the quadratic equation to solve this problem by equating the quadratic to $0$ to find the roots or the zeros:

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 $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{11}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{\left(- 3\right)} \pm \sqrt{{\textcolor{b l u e}{\left(- 3\right)}}^{2} - \left(4 \cdot \textcolor{red}{- 1} \cdot \textcolor{g r e e n}{11}\right)}}{2 \cdot \textcolor{red}{- 1}}$

$x = \frac{\textcolor{b l u e}{\left(- 3\right)} \pm \sqrt{9 - \left(- 44\right)}}{- 2}$

$x = \frac{\textcolor{b l u e}{\left(- 3\right)} \pm \sqrt{9 + 44}}{- 2}$

$x = \frac{- 3 \pm \sqrt{53}}{- 2}$

$x = \frac{3 \pm \sqrt{53}}{2}$