# How do you find the zeros, real and imaginary, of y=-6x^2-12x+9 using the quadratic formula?

Jul 31, 2017

$x = \frac{- 2 \pm \sqrt{10}}{2}$

#### Explanation:

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

Here, $a = - 6$, $b = - 12$, $c = 9$
So,

$x = \frac{12 \pm \sqrt{144 + 216}}{- 12}$
$\Rightarrow$ $x = \frac{- 2 \pm \sqrt{10}}{2}$

Or in decimal form,

$x = 0.58113883008418966599944677221636$
or, $x = - 2.5811388300841896659994467722164$