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

Oct 17, 2016

The roots (both real) are $x = 3 + \sqrt{3}$ and $x = 3 - \sqrt{3}$.

#### Explanation:

A quadratic function in standard form ($y = a {x}^{2} + b x + c$) can be solved using its related quadratic equation ($a {x}^{2} + b x + c = 0$) and the quadratic formula. The quadratic formula is:

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

For the quadratic function $y = - {x}^{2} - 6 x - 6$, $a = - 1$, $b = - 6$, and $c = - 6$. Substitute these values into the quadratic formula and simplify following the Order of Operations.

$x = \frac{- \left(- 6\right) \pm \sqrt{{\left(- 6\right)}^{2} - 4 \left(- 1\right) \left(- 6\right)}}{\left(2\right) \left(- 1\right)}$
$x = \frac{6 \pm \sqrt{36 - 24}}{-} 2$
$x = \frac{6 \pm \sqrt{12}}{-} 2$
$x = \frac{- 6 \pm 2 \sqrt{3}}{-} 2$
$x = 3 \pm \sqrt{3}$

So, the roots of $y = - {x}^{2} - 6 x - 6$ are $x = 3 + \sqrt{3}$ and $x = 3 - \sqrt{3}$.