# How do you find the roots, real and imaginary, of y=-(2x+2)^2+6x^2+3x-12  using the quadratic formula?

Jan 5, 2018

See a solution process below:

#### Explanation:

First, we need to expand the ${\left(2 x + 2\right)}^{2}$ term using this rule for quadratics:

${\left(\textcolor{red}{x} + \textcolor{b l u e}{y}\right)}^{2} = \left(\textcolor{red}{x} + \textcolor{b l u e}{y}\right) \left(\textcolor{red}{x} + \textcolor{b l u e}{y}\right) = {\textcolor{red}{x}}^{2} + 2 \textcolor{red}{x} \textcolor{b l u e}{y} + {\textcolor{b l u e}{y}}^{2}$

${\left(\textcolor{red}{2 x} + \textcolor{b l u e}{2}\right)}^{2} = \left(\textcolor{red}{2 x} + \textcolor{b l u e}{2}\right) \left(\textcolor{red}{2 x} + \textcolor{b l u e}{2}\right) =$

${\left(\textcolor{red}{2 x}\right)}^{2} + \left(2 \times \textcolor{red}{2 x} \times \textcolor{b l u e}{2}\right) + {\textcolor{b l u e}{2}}^{2} = 4 {x}^{2} + 8 x + 4$

Substituting and simplifying gives:

$y = - {\left(2 x + 2\right)}^{2} + 6 {x}^{2} + 3 x - 12$

$y = - \left(4 {x}^{2} + 8 x + 4\right) + 6 {x}^{2} + 3 x - 12$

$y = - 4 {x}^{2} - 8 x - 4 + 6 {x}^{2} + 3 x - 12$

$y = - 4 {x}^{2} + 6 {x}^{2} - 8 x + 3 x - 4 - 12$

$y = \left(- 4 + 6\right) {x}^{2} + \left(- 8 + 3\right) x + \left(- 4 - 12\right)$

$y = 2 {x}^{2} + \left(- 5\right) x + \left(- 16\right)$

$y = 2 {x}^{2} - 5 x - 16$

We can now use the quadratic equation to solve this problem:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{2}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 5}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 16}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{- 5} \pm \sqrt{{\textcolor{b l u e}{- 5}}^{2} - \left(4 \cdot \textcolor{red}{2} \cdot \textcolor{g r e e n}{- 16}\right)}}{2 \cdot \textcolor{red}{2}}$

$x = \frac{5 \pm \sqrt{25 - \left(- 128\right)}}{4}$

$x = \frac{5 \pm \sqrt{25 + 128}}{4}$

$x = \frac{5 \pm \sqrt{153}}{4}$

$x = \frac{5 \pm \sqrt{9 \times 17}}{4}$

$x = \frac{5 \pm \sqrt{9} \sqrt{17}}{4}$

$x = \frac{5 \pm 3 \sqrt{17}}{4}$