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

$x = \frac{9 + \sqrt{113}}{- 2} \cong 9.8 \mathmr{and} x = \frac{9 - \sqrt{113}}{- 2} \cong - 0.8$

#### Explanation:

Let's first simplify this so that we can have it in the form of $y = a {x}^{2} + b x + c$:

$y = - 3 {x}^{2} + - x + 2 {\left(x - 2\right)}^{2}$

$y = - 3 {x}^{2} - x + 2 \left({x}^{2} - 4 x + 4\right)$

$y = - 3 {x}^{2} - x + 2 {x}^{2} - 8 x + 8$

$y = - {x}^{2} - 9 x + 8$

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

and we have $a = - 1 , b = - 9 , c = 8$

$x = \frac{9 \setminus \pm \sqrt{{\left(- 9\right)}^{2} - 4 \left(- 1\right) \left(8\right)}}{2 \left(- 1\right)}$

$x = \frac{9 \setminus \pm \sqrt{81 + 32}}{- 2}$

$x = \frac{9 \setminus \pm \sqrt{113}}{- 2}$

$x = \frac{9 + \sqrt{113}}{- 2} \cong 9.8 \mathmr{and} x = \frac{9 - \sqrt{113}}{- 2} \cong - 0.8$