# How do you find the zeros, real and imaginary, of y=-x^2 -16x +8 using the quadratic formula?

Dec 23, 2017

The zeros of the equation are $8 \pm 6 \sqrt{2}$.

#### Explanation:

$y = - {x}^{2} - 16 x + 8$

We can solve for any zero(s) of a function with the quadratic formula, stated here: $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Our equation is written in standard form, or $y = a {x}^{2} + b x + c$.

We know that $a = - 1$, $b = - 16$, and $c = 8$, so we plug in those values into the quadratic formula and solve for $x$:

$x = \frac{- \left(- 16\right) \pm \sqrt{{\left(- 16\right)}^{2} - 4 \left(- 1\right) \left(8\right)}}{2 \left(- 1\right)}$

$x = \frac{16 \pm \sqrt{256 - 4 \left(- 8\right)}}{-} 2$

$x = \frac{16 \pm \sqrt{256 + 32}}{2}$

$x = \frac{16 \pm \sqrt{288}}{2}$

$x = \frac{16 \pm 12 \sqrt{2}}{2}$

$x = 8 \pm 6 \sqrt{2}$

The zeros of the equation are $8 \pm 6 \sqrt{2}$.

Hope this helps!