# How do you find the roots, real and imaginary, of y=-2(x - 2)^2+4x^2-3x + 2 using the quadratic formula?

May 27, 2017

$x = \frac{- 5 - \sqrt{73}}{4} , \frac{- 5 + \sqrt{73}}{4}$

#### Explanation:

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

Let's expand the parentheses:

$R i g h t a r r o w y = - 2 \left({x}^{2} - 4 x + 4\right) + 4 {x}^{2} - 3 x + 2$

$R i g h t a r r o w y = - 2 {x}^{2} + 8 x - 8 + 4 {x}^{2} - 3 x + 2$

Then, let's collect all like terms:

$R i g h t a r r o w y = 2 {x}^{2} + 5 x - 6$

$R i g h t a r r o w x = \frac{- 5 \pm \sqrt{{5}^{2} - 4 \left(2\right) \left(- 6\right)}}{2 \left(2\right)}$
$R i g h t a r r o w x = \frac{- 5 \pm \sqrt{25 + 48}}{4}$
$\therefore x = \frac{- 5 \pm \sqrt{73}}{4}$
Therefore, the solutions to the equation are $x = \frac{- 5 \pm - \left(73\right)}{4}$ and $x = \frac{- 5 + \sqrt{73}}{4}$.