# How do you find the roots, real and imaginary, of y=(x+4)(x+1)-3x^2+2 using the quadratic formula?

Jul 14, 2016

$x = \frac{5}{4} \pm \frac{\sqrt{73}}{4}$

#### Explanation:

Multiply out the bracket:

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

$\therefore y = - 2 {x}^{2} + 5 x + 6$

For quadratic $y = a {x}^{2} + b x + c$ the roots occur at $y = 0$ and are found with the quadratic formula. As it is a polynomial of order 2, we expect to get two roots.

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

$x = \frac{- 5 \pm \sqrt{25 - 4 \left(- 2\right) \left(6\right)}}{2 \left(- 2\right)} = \frac{- 5 \pm \sqrt{73}}{- 4}$

$\therefore x = \frac{5}{4} \pm \frac{\sqrt{73}}{4}$