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

Nov 9, 2017

Expand the brackets and use the Formula

$x = \frac{1}{4} \left(13 + \sqrt{95} i\right)$ or $x = \frac{1}{4} \left(13 - \sqrt{95} i\right)$

#### Explanation:

Firstly, expand the brackets and collect like terms.

$y = - \left({x}^{2} - 12 x + 36\right) - {x}^{2} + x + 3$
$y = - {x}^{2} + 12 x - 36 - {x}^{2} + x + 3$
$y = - 2 {x}^{2} + 13 x - 33$

Now we sub into the quadratic formula $a = - 2 , b = 13 , c = - 33$

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

x=(-13+-sqrt(13^2-4(-2)(-33)))/(2(-2)

$x = \frac{- 13 \pm \sqrt{169 - 264}}{- 4}$

$x = \frac{13 \pm \sqrt{- 95}}{4}$

Ok, we can see here our roots will be imaginary, but that's ok.

$x = \frac{13 \pm \sqrt{95} i}{4}$

$x = \frac{1}{4} \left(13 + \sqrt{95} i\right)$ or $x = \frac{1}{4} \left(13 - \sqrt{95} i\right)$