# How do you find the zeros, real and imaginary, of y= -9x^2-28x-3 using the quadratic formula?

Jul 26, 2018

The zeros are $x = - 3 , - \frac{1}{9}$.

#### Explanation:

$y = - 9 {x}^{2} - 28 x - 3$

This equation is in standard quadratic form, or $y = \textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{m a \ge n t a}{c}$, where $\textcolor{red}{a = - 9}$, $\textcolor{b l u e}{b = - 28}$, and $\textcolor{m a \ge n t a}{c = - 3}$.

The quadratic formula is $x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{m a \ge n t a}{c}}}{2 \textcolor{red}{a}}$.

Plug in the values and solve:
$x = \frac{- \left(\textcolor{b l u e}{- 28}\right) \pm \sqrt{{\left(\textcolor{b l u e}{- 28}\right)}^{2} - 4 \left(\textcolor{red}{- 9}\right) \left(\textcolor{m a \ge n t a}{- 3}\right)}}{2 \left(\textcolor{red}{- 9}\right)}$:

$x = \frac{28 \pm \sqrt{784 - 108}}{- 18}$

$x = \frac{28 \pm \sqrt{676}}{- 18}$

$x = \frac{28 \pm 26}{- 18}$

$x = \frac{28 + 26}{- 18}$, $\frac{28 - 26}{- 18}$

$x = \frac{54}{-} 18$, $\frac{2}{-} 18$

$x = - 3 , - \frac{1}{9}$

Hope this helps!