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

Aug 25, 2017

$x = \frac{7}{2} \pm \frac{\sqrt{85}}{2}$

#### Explanation:

By roots, I will assume you mean zeros.

First expand out the quadratic, then combine terms to bring it into standard form:

$2 {x}^{2} - x - {\left(x + 3\right)}^{2} = 2 {x}^{2} - x - \left({x}^{2} + 6 x + 9\right)$

$\textcolor{w h i t e}{2 {x}^{2} - x - {\left(x + 3\right)}^{2}} = 2 {x}^{2} - x - {x}^{2} - 6 x - 9$

$\textcolor{w h i t e}{2 {x}^{2} - x - {\left(x + 3\right)}^{2}} = {x}^{2} - 7 x - 9$

This is in the form:

$a {x}^{2} + b x + c$

with $a = 1$, $b = - 7$ and $c = - 9$

It has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(\textcolor{b l u e}{- 7}\right)}^{2} - 4 \left(\textcolor{b l u e}{1}\right) \left(\textcolor{b l u e}{- 9}\right) = 49 + 36 = 85$

Since $\Delta > 0$, this quadratic has two distinct real zeros, but since $85$ is not a perfect square, those zeros are irrational.

We can use the quadratic formula to find them:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{7 \pm \sqrt{85}}{2}$

$\textcolor{w h i t e}{x} = \frac{7}{2} \pm \frac{\sqrt{85}}{2}$