# How do you find the zeros, real and imaginary, of y= -7x^2-3x-16 using the quadratic formula?

Plug the coefficients into the quadratic formula and get the rather messy $x = - \frac{3}{14} + \frac{439}{14} i$ and $x = - \frac{3}{14} - \frac{439}{14} i$ as a solution

#### Explanation:

We'll be using the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Let's substitute in:

$x = \frac{3 \pm \sqrt{- {3}^{2} - 4 \left(- 7\right) \left(- 16\right)}}{2 \left(- 7\right)}$

and simplify:

$x = \frac{3 \pm \sqrt{9 - 448}}{- 14}$
$x = \frac{3 \pm \sqrt{- 439}}{- 14}$

So the roots are:

$x = - \frac{3}{14} + \frac{\sqrt{- 439}}{14}$ and $x = - \frac{3}{14} - \frac{\sqrt{- 439}}{14}$

A negative under the square root makes that number imaginary, so we use the following symbology:

$x = - \frac{3}{14} + \frac{439}{14} i$ and $x = - \frac{3}{14} - \frac{439}{14} i$