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

Sep 4, 2017

$x = - 6 + 5 \sqrt{2} \text{ or } x = - 6 - 5 \sqrt{2}$

#### Explanation:

Need to expand out the brackets:

$y = 2 \left({x}^{2} + 2 x + 1\right) - \left({x}^{2} - 8 x + 16\right)$

$y = {x}^{2} + 12 x - 14$

The roots of a quadratic of the form

$a {x}^{2} + b x + c$

given by

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

In this case $a = 1 , b = 12 , c = - 14$

$x = \frac{- 12 \pm \sqrt{144 - 4 \left(1\right) \left(- 14\right)}}{2} = \frac{- 12 \pm \sqrt{144 + 56}}{2}$

$x = \frac{- 12 \pm \sqrt{200}}{2}$

$x = \frac{- 12 \pm \sqrt{100} \sqrt{2}}{2} = \frac{- 12 \pm 10 \sqrt{2}}{2}$

$\therefore x = - 6 + 5 \sqrt{2} \text{ or } x = - 6 - 5 \sqrt{2}$