# How do you find the roots, real and imaginary, of y= -x^2-3x-2(x-1)^2  using the quadratic formula?

Jan 26, 2018

Imaginary Roots , $\frac{- 1 \pm \sqrt{23} i}{- 6}$

#### Explanation:

Simply this equation to get : $y = a {x}^{2} + b x + c$

$y = - {x}^{2} - 3 x - 2 \left({x}^{2} - 2 x + 1\right)$
$y = - {x}^{2} - 3 x - 2 {x}^{2} + 4 x - 2$

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

comparing with: $y = a {x}^{2} + b x + c$

we get $a = - 3$ , $b = 1$, $c = - 2$

$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
(-1+-sqrt(1^2-4(-3)(-2)))/(2(-3)
answer = $\frac{- 1 \pm \sqrt{23} i}{- 6}$