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

Dec 27, 2017

${x}_{1} = 0$
${x}_{2} = - \frac{1}{3}$

#### Explanation:

For the function in the form of:
$a {x}^{2} + b x + c = 0$

${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$y = 5 {x}^{2} - x - 2 x \left(x - 1\right) = 5 {x}^{2} - x - 2 {x}^{2} + 2 x = 3 {x}^{2} + x$

Let $y = 0$:
$3 {x}^{2} + 1 x + 0 = 0$
where:
$a = 3$
$b = 1$
$c = 0$
$\implies$
${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \cdot 3 \cdot 0}}{2 \cdot 3} =$
$= \frac{- 1 \pm \sqrt{1}}{6} = \frac{- 1 \pm 1}{6}$
$\implies$
${x}_{1} = \frac{- 1 + 1}{6} = \frac{0}{6} = 0$
${x}_{2} = \frac{- 1 - 1}{6} = \frac{- 2}{6} = - \frac{1}{3}$