# How do you find the roots, real and imaginary, of y=(x-2)(-5x+1)-x using the quadratic formula?

Jun 5, 2016

$x = 1 \pm \frac{\sqrt{60}}{10}$

#### Explanation:

First convert the given equation into standard form:

$y = \left(x - 2\right) \left(- 5 x + 1\right) - x$

$\textcolor{w h i t e}{\text{XXX}} \Rightarrow y = - 5 {x}^{2} + 10 x - 2$

$\textcolor{w h i t e}{\text{XXX}} y = a {x}^{2} + b x + c$
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- 10 \pm \sqrt{{10}^{2} - 4 \left(- 5\right) \left(- 2\right)}}{2 \left(- 5\right)}$
$\textcolor{w h i t e}{\text{XXX}} x = 1 \pm \frac{\sqrt{60}}{10}$