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

Jun 8, 2016

$\frac{- 7 \pm \sqrt{41}}{2}$

Explanation:

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

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

= $- {x}^{2} - 7 x - 2$

As according to quadratic formula roots of $a {x}^{2} + b x + c = 0$ are

$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$, roots of $- {x}^{2} - 7 x - 2$ are

$\frac{- \left(- 7\right) \pm \sqrt{{\left(- 7\right)}^{2} - 4 \left(- 1\right) \left(- 2\right)}}{2 \left(- 1\right)}$

= $\frac{7 \pm \sqrt{49 - 8}}{- 2}$

= $\frac{- 7 \pm \sqrt{41}}{2}$