# How do you find the zeros of  y = -x^2 - 12x + 7  using the quadratic formula?

Nov 28, 2015

The zeros are at
$x = - 6 + \sqrt{43}$ and $x = - 6 - \sqrt{43}$

#### Explanation:

$a {x}^{2} + b x + c = 0 \implies x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

We are looking for the values of $x$ at which

$- {x}^{2} - 12 x + 7 = 0$

Then we can apply the quadratic formula with
$a = - 1 , b = - 12 , c = 7$
to obtain

$x = \frac{- \left(- 12\right) \pm \sqrt{{\left(- 12\right)}^{2} - 4 \left(- 1\right) \left(7\right)}}{2 \left(- 1\right)}$

$= \frac{12 \pm \sqrt{144 + 28}}{- 2}$

$= - \frac{12 \pm \sqrt{172}}{2}$

$= - \frac{12 \pm 2 \sqrt{43}}{2}$

$= - 6 \pm \sqrt{43}$

Thus the zeros are at
$x = - 6 + \sqrt{43}$ and $x = - 6 - \sqrt{43}$