# How do you find the roots, real and imaginary, of y= -x^2 -x-6-(x- 3 )^2  using the quadratic formula?

Mar 15, 2018

$x = \frac{- 5 + i \sqrt{95}}{-} 4$ and
$x = \frac{- 5 - i \sqrt{95}}{-} 4$

#### Explanation:

First, you want to fully expand the equation then combine all like terms.
$- {x}^{2} - x - 6 - {\left(x - 3\right)}^{2} = 0$

$- {x}^{2} - x - 6 - {x}^{2} + 6 x - 9 = 0$

$- 2 {x}^{2} + 5 x - 15 = 0$

Now the equation is simplified to be able to be used in the quadratic formula.

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

$a {x}^{2} + b x + c = 0$

$\left(- 2\right) {x}^{2} + \left(5\right) x + \left(- 15\right) = 0$

$x = \frac{- 5 \pm \sqrt{{5}^{2} - \left(4 \cdot - 2 \cdot - 15\right)}}{2 \cdot - 2}$

$x = \frac{- 5 \pm \sqrt{- 95}}{-} 4$

$x = \frac{- 5 \pm i \sqrt{95}}{-} 4$