How do you find the zeros, real and imaginary, of y= -3x^2-2x-18 using the quadratic formula?

Aug 21, 2017

See a solution process below:

Explanation:

To find the zeroes we set $y$ to $0$ giving:

$0 = - 3 {x}^{2} - 2 x - 18$

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:

$\textcolor{red}{- 3}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 2}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 18}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{\left(- 2\right)} \pm \sqrt{{\textcolor{b l u e}{\left(- 2\right)}}^{2} - \left(4 \cdot \textcolor{red}{- 3} \cdot \textcolor{g r e e n}{- 18}\right)}}{2 \cdot \textcolor{red}{- 3}}$

$x = \frac{2 \pm \sqrt{4 - \left(216\right)}}{- 6}$

$x = \frac{2 \pm \sqrt{- 212}}{- 6}$

$x = \frac{2 \pm \sqrt{4 \cdot - 53}}{- 6}$

$x = \frac{2 \pm \sqrt{4} \sqrt{- 53}}{- 6}$

$x = \frac{2 \pm 2 \sqrt{- 53}}{- 6}$

$x = \frac{1 \pm 1 \sqrt{- 53}}{- 3}$

$x = \frac{1 \pm \sqrt{- 53}}{- 3}$