Question

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

Answer 1

There are two real roots: `x=-6` and `x=5/2`. There are no imaginary roots.

Explanation:

First, expand the original equation to `y=ax^2+bx+c` form.

`y=-2x^2+5x-12x+30=-2x^2-7x+30`

Next plug the coefficients into the quadratic formula, giving you

`x=(7+-sqrt(49-4(-2)(30)))/(2(-2))`
`=(7+-sqrt(289))/(-4)`
`=(7+-17)/(-4)`

So

`x=24/-4=-6` or `x=(-10)/-4=5/2`