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

1 Answer
Mar 14, 2016

Answer:

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#