# How do you find the zeros, real and imaginary, of y=- 5x^2-15x -3 using the quadratic formula?

Aug 4, 2018

$x = \frac{15 + \sqrt{165}}{-} 10$ and $x = \frac{15 - \sqrt{165}}{-} 10$

#### Explanation:

$y = - 5 {x}^{2} - 15 x - 3$

The quadratic formula is $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

We know that: $a = - 5$, $b = - 15$, and $c = - 3$ based on the equation, so let's plug them into the formula:

$x = \frac{- \left(- 15\right) \pm \sqrt{{\left(- 15\right)}^{2} - 4 \left(- 5\right) \left(- 3\right)}}{2 \left(- 5\right)}$

$x = \frac{15 \pm \sqrt{225 - 60}}{-} 10$

$x = \frac{15 \pm \sqrt{165}}{-} 10$

Hope this helps!