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

$x = \frac{2 i \sqrt{3}}{3} , - \frac{2 i \sqrt{3}}{3}$

#### Explanation:

We can first expand the bracket:

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

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

and now distribute the 4:

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

$y = 2 {x}^{2} + 4 x + {x}^{2} - 4 x + 4$

combine terms:

$y = 3 {x}^{2} + 4$

and now see that we can use the quadratic formula, with values $a = 3 , b = 0 , c = 4$

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

$x = \frac{0 \setminus \pm \sqrt{{0}^{2} - 4 \left(3\right) \left(4\right)}}{2 \left(3\right)}$

$x = \frac{\setminus \pm \sqrt{- 48}}{6}$

$x = \frac{\setminus \pm 4 i \sqrt{3}}{6} = \pm \frac{2 i \sqrt{3}}{3}$