# How do you find the roots, real and imaginary, of y=x(2x-1)-(3x-1)^2  using the quadratic formula?

Aug 3, 2016

$x = \frac{5}{14} \pm \frac{\sqrt{3}}{14} i$

#### Explanation:

Let's multiply out the brackets and get ourselves some quadratic terms:

$y = 2 {x}^{2} - x - 9 {x}^{2} + 6 x - 1 = - 7 {x}^{2} + 5 x - 1$

The quadratic formula for polynomial of the form $a {x}^{2} + b x + c = 0$ is given by:

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

$a = - 7 , b = 5 , c = - 1$

$\therefore x = \frac{- 5 \pm \sqrt{25 - 4 \left(- 7\right) \left(- 1\right)}}{2 \left(- 7\right)}$

$x = \frac{- 5 \pm \sqrt{- 3}}{- 14}$

Recall that ${i}^{2} = - 1$ so we can rewrite $\sqrt{- 3}$ as $\sqrt{3 {i}^{2}} = \sqrt{3} i$

$\therefore x = \frac{5}{14} \pm \frac{\sqrt{3}}{14} i$