# How do you find the roots, real and imaginary, of y= 5x^2 - 13x + 4+x(x-1)  using the quadratic formula?

Aug 14, 2017

See below.

#### Explanation:

Solve

$5 {x}^{2} - 13 x + 4 + x \left(x - 1\right) = 6 {x}^{2} - 14 x + 4 = 2 \left(3 {x}^{2} - 7 x + 2\right) = 0$

Which is equivalent to

$3 {x}^{2} - 7 x + 2 = 0$

We were told to use the quadratic formula, so

$x = \frac{- \left(- 7\right) \pm \sqrt{{\left(- 7\right)}^{2} - 4 \left(3\right) \left(2\right)}}{2 \left(3\right)}$

$= \frac{7 \pm \sqrt{25}}{6} = \frac{7 \pm 5}{6}$

$x = \frac{12}{6} = 2$ or $x = \frac{2}{6} = \frac{1}{3}$