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

May 21, 2018

$x = \frac{5 \pm i \cdot \sqrt{2}}{3}$

#### Explanation:

First, square $\left(x - 2\right)$ and combine like terms.
$y = 3 {\left(x - 2\right)}^{2} + 2 x - 3$
$y = 3 \left(x - 2\right) \left(x - 2\right) + 2 x - 3$
$y = 3 \left({x}^{2} - 4 x + 4\right) + 2 x - 3$
$y = 3 {x}^{2} - 12 x + 12 + 2 x - 3$
$y = 3 {x}^{2} - 10 x + 9$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- \left(- 10\right) \pm \sqrt{{\left(- 10\right)}^{2} - 4 \left(3\right) \left(9\right)}}{2 \left(3\right)}$
$x = \frac{10 \pm \sqrt{100 - 108}}{6}$
$x = \frac{10 \pm \sqrt{- 8}}{6}$
$x = \frac{10 \pm 2 i \cdot \sqrt{2}}{6}$
$x = \frac{5 \pm i \cdot \sqrt{2}}{3}$