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

Aug 15, 2017

$x = - \frac{1 + \sqrt{13}}{3} ,$$- \frac{1 - \sqrt{13}}{3}$

Refer to the explanation for the process.

#### Explanation:

Given:

$y = \left(2 x - 4\right) \left(x - 1\right) - 5 {x}^{2} + 4 x$

FOIL $\left(2 x - 4\right) \left(x - 1\right)$.

$y = \left[2 {x}^{2} - 2 x - 4 x + 4\right] - 5 {x}^{2} + 4 x$

Simplify.

$y = \left[2 {x}^{2} - 6 x + 4\right] - 5 {x}^{2} + 4 x$

Gather like terms.

$y = \left(2 {x}^{2} - 5 {x}^{2}\right) + \left(- 6 x + 4 x\right) + 4$

Combine like terms.

$y = - 3 {x}^{2} - 2 x + 4$ $\leftarrow$ quadratic equation standard form:

$y = a {x}^{2} + b x + c$,

where:

$a = - 3$, $b = - 2$, and $c = 4$.

Substitute $0$ for $y$. Solve for $x$.

$0 = - 3 {x}^{2} - 2 x + 4$

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

Plug in the known values.

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

Simplify.

$x = \frac{2 \pm \sqrt{4 + 48}}{- 6}$

$x = \frac{2 \pm \sqrt{52}}{- 6}$

Prime factorize $52$.

$x = \frac{2 \pm \sqrt{2 \times 2 \times 13}}{- 6}$

Simplify.

$\frac{2 \pm 2 \sqrt{13}}{- 6}$

Simplify.

(color(red)cancel(color(black)(2^1))+-(color(red)cancel(color(black)(2^1))sqrt13))/(-(color(red)cancel(color(black)(6^3)))

$x = \frac{1 \pm \sqrt{3}}{-} 3$

Roots

$x = - \frac{1 + \sqrt{13}}{3} ,$$- \frac{1 - \sqrt{13}}{3}$