# How do you find the roots, real and imaginary, of y=(x +4)^2+4x^2-3x + 2 using the quadratic formula?

Sep 3, 2017

See a solution process below:

#### Explanation:

First, we need to write the right side of the equation in standard form:
$y = {\left(x + 4\right)}^{2} + 4 {x}^{2} - 3 x + 2$

$y = {x}^{2} + 8 x + 16 + 4 {x}^{2} - 3 x + 2$

$y = 1 {x}^{2} + 4 {x}^{2} + 8 x - 3 x + 16 + 2$

$y = \left(1 + 4\right) {x}^{2} + \left(8 - 3\right) x + \left(16 + 2\right)$

$y = 5 {x}^{2} + 5 x + 18$

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:

$\textcolor{red}{5}$ for $\textcolor{red}{a}$

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

$\textcolor{g r e e n}{18}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{5} \pm \sqrt{{\textcolor{b l u e}{5}}^{2} - \left(4 \cdot \textcolor{red}{5} \cdot \textcolor{g r e e n}{18}\right)}}{2 \cdot \textcolor{red}{5}}$

$x = \frac{- \textcolor{b l u e}{5} \pm \sqrt{25 - 360}}{10}$

$x = \frac{- 5 \pm \sqrt{- 335}}{10}$