# How do you find the roots, real and imaginary, of y=12x^2 -7x +28-(x-2)^2  using the quadratic formula?

##### 1 Answer
Apr 13, 2018

$x = \frac{3}{22} \pm \frac{\sqrt{1047}}{22} i$

#### Explanation:

The quadratic formula states that, for a quadratic in the form

$y = \textcolor{red}{a} {x}^{2} + \textcolor{g r e e n}{b} x + \textcolor{b l u e}{c} \implies x = \frac{- \textcolor{g r e e n}{b} \pm \sqrt{{\textcolor{g r e e n}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{b l u e}{c}}}{2 \textcolor{red}{a}}$

We need to put our given quadratic in the form $y = \textcolor{red}{a} {x}^{2} + \textcolor{g r e e n}{b} x + \textcolor{b l u e}{c}$

$y = 12 {x}^{2} - 7 x + 28 - {\left(x - 2\right)}^{2}$

$= 12 {x}^{2} - 7 x + 28 - \left({x}^{2} - 4 x + 4\right)$

$= \textcolor{red}{11} {x}^{2} \textcolor{g r e e n}{- 3} x + \textcolor{b l u e}{24}$

From the quadratic formula:

$x = \frac{- \textcolor{g r e e n}{\left(- 3\right)} \pm \sqrt{{\textcolor{g r e e n}{\left(- 3\right)}}^{2} - 4 \times \textcolor{red}{11} \times \textcolor{b l u e}{24}}}{2 \times \textcolor{red}{11}}$

$= \frac{3 \pm \sqrt{- 1047}}{22}$

We can separate $\sqrt{- 1047}$ into $\sqrt{1047} \times \sqrt{- 1}$. Since $i = \sqrt{- 1}$:

$x = \frac{3 \pm \sqrt{1047} i}{22}$

$x = \frac{3}{22} \pm \frac{\sqrt{1047}}{22} i$