# How do you find the zeros of  y = -2x^2 - 3x +12  using the quadratic formula?

Dec 2, 2015

Set $a$, $b$, and $c$ appropriately and apply the quadratic formula to obtain the zeros at $x = \frac{- 3 + \sqrt{105}}{4}$ and $x = \frac{- 3 - \sqrt{105}}{4}$

#### Explanation:

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

In this case, we have
$a = - 2$
$b = - 3$
$c = 12$

Thus the zeros, that is, the solutions to $- 2 {x}^{2} - 3 x + 12 = 0$ are

$x = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(- 2\right) \left(12\right)}}{2 \left(- 2\right)}$

$= \frac{3 \pm \sqrt{9 + 96}}{- 4}$

$= - \frac{3 \pm \sqrt{105}}{4}$

$= \frac{- 3 \pm \sqrt{105}}{4}$

So
$x = \frac{- 3 + \sqrt{105}}{4}$
or
$x = \frac{- 3 - \sqrt{105}}{4}$