# How do you find the roots, real and imaginary, of y= 2x^2 - 3x + 4+ 4(-x-1)^2  using the quadratic formula?

Jan 4, 2018

#### Answer:

no real solutions

#### Explanation:

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

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

$\therefore y = 2 {x}^{2} - 3 x + 4 + 4 {x}^{2} + 8 x + 4$

$\therefore y = 4 {x}^{2} + 5 x + 8$

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

$\therefore y = \frac{- \left(5\right) \pm \sqrt{{\left(5\right)}^{2} - 4 \left(6\right)} \left(8\right)}{2 \left(6\right)}$

$\therefore y = \frac{- 5 \pm \sqrt{- 167}}{12}$

no real solutions.