# How do you find the roots, real and imaginary, of y= 2x^2 - 2x -1 - (x-7)^2  using the quadratic formula?

Nov 13, 2017

See a solution process below:

#### Explanation:

First, we need to put the equation in standard form by:
1. Expanding the terms in parenthesis
2. Grouping like terms
3. Combining like terms

$y = 2 {x}^{2} - 2 x - 1 - {\left(x - 7\right)}^{2}$

$y = 2 {x}^{2} - 2 x - 1 - \left({x}^{2} - 7 x - 7 x + 49\right)$

$y = 2 {x}^{2} - 2 x - 1 - {x}^{2} + 7 x + 7 x - 49$

$y = 2 {x}^{2} - {x}^{2} - 2 x + 7 x + 7 x - 1 - 49$

$y = 2 {x}^{2} - 1 {x}^{2} - 2 x + 7 x + 7 x - 1 - 49$

$y = \left(2 - 1\right) {x}^{2} + \left(- 2 + 7 + 7\right) x + \left(- 1 - 49\right)$

$y = 1 {x}^{2} + 12 x + \left(- 50\right)$

$y = 1 {x}^{2} + 12 x - 50$

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:

$\textcolor{red}{1}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{12}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 50}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{12} \pm \sqrt{{\textcolor{b l u e}{12}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{- 50}\right)}}{2 \cdot \textcolor{red}{1}}$

$x = \frac{- 12 \pm \sqrt{144 - \left(- 200\right)}}{2}$

$x = \frac{- 12 \pm \sqrt{144 + 200}}{2}$

$x = \frac{- 12 \pm \sqrt{344}}{2}$

$x = \frac{- 12 \pm \sqrt{4 \cdot 86}}{2}$

$x = \frac{- 12 \pm \sqrt{4} \sqrt{86}}{2}$

$x = \frac{- 12 \pm 2 \sqrt{86}}{2}$

$x = - 6 \pm \sqrt{86}$