# How do you find the roots, real and imaginary, of y=(3x-4)(x+2)-x^2-x  using the quadratic formula?

$\therefore x = \frac{- 1 + \sqrt{33}}{4} \cong 1.186$

$\therefore x = \frac{- 1 - \sqrt{33}}{4} \cong - 1.686$

There are no imaginary roots.

#### Explanation:

Let's first simplify:

$y = \left(3 x - 4\right) \left(x + 2\right) - {x}^{2} - x$

$y = 3 {x}^{2} + 2 x - 8 - {x}^{2} - x$

$y = 2 {x}^{2} + x - 8$

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

with $a = 2 , b = 1 , c = - 8$

$x = \frac{- 1 \setminus \pm \sqrt{{1}^{2} - 4 \left(2\right) \left(- 8\right)}}{2 \left(2\right)}$

$x = \frac{- 1 \setminus \pm \sqrt{{1}^{2} + 32}}{4}$

$x = \frac{- 1 \setminus \pm \sqrt{33}}{4}$

$\therefore x = \frac{- 1 + \sqrt{33}}{4} \cong 1.186$

$\therefore x = \frac{- 1 - \sqrt{33}}{4} \cong - 1.686$

There are no imaginary roots.