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

Dec 29, 2015

$y = 2 {\left(x + 3\right)}^{2} - {\left(x - 3\right)}^{2}$
$\implies y = 2 \left({x}^{2} + 6 x + 9\right) - \left({x}^{2} - 6 x + 9\right)$
$\implies y = 2 {x}^{2} + 12 x + 18 - {x}^{2} + 6 x - 9$
$\implies y = {x}^{2} + 18 x + 9$
For finding roots  becomes 0. implies x^2+18x+9=0

If $a {x}^{2} + b x + c = 0$ then $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
Here $a = 1$ $b = 18$ and $c = 9$

$\implies x = \frac{- 18 \pm \sqrt{{18}^{2} - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

$\implies x = \frac{- 18 \pm \sqrt{324 - 36}}{2}$

$\implies x = \frac{- 18 \pm \sqrt{324 - 36}}{2}$

$\implies x = \frac{- 18 \pm \sqrt{288}}{2}$

$\implies x = \frac{- 18 \pm \sqrt{288}}{2}$

$\implies x = \frac{- 18 \pm 12 \sqrt{2}}{2}$

$\implies x = \frac{2 \left(- 9 \pm 6 \sqrt{2}\right)}{2}$

$\implies x = \left(- 9 \pm 6 \sqrt{2}\right)$