How do you find the roots, real and imaginary, of y=(x+6)(-2x+5) using the quadratic formula?

Mar 14, 2016

There are two real roots: $x = - 6$ and $x = \frac{5}{2}$. There are no imaginary roots.

Explanation:

First, expand the original equation to $y = a {x}^{2} + b x + c$ form.

$y = - 2 {x}^{2} + 5 x - 12 x + 30 = - 2 {x}^{2} - 7 x + 30$

Next plug the coefficients into the quadratic formula, giving you

$x = \frac{7 \pm \sqrt{49 - 4 \left(- 2\right) \left(30\right)}}{2 \left(- 2\right)}$
$= \frac{7 \pm \sqrt{289}}{- 4}$
$= \frac{7 \pm 17}{- 4}$

So

$x = \frac{24}{-} 4 = - 6$ or $x = \frac{- 10}{-} 4 = \frac{5}{2}$