How do you find the roots, real and imaginary, of y= -x^2 -x-6-(x- 3 )^2 using the quadratic formula?

1 Answer
Mar 15, 2018

x=(-5+isqrt95)/-4 and
x=(-5-isqrt95)/-4

Explanation:

First, you want to fully expand the equation then combine all like terms.
-x^2-x-6-(x-3)^2=0

-x^2-x-6-x^2+6x-9=0

-2x^2+5x-15=0

Now the equation is simplified to be able to be used in the quadratic formula.

x=(-b+-sqrt(b^2-4ac))/(2a)

ax^2+bx+c=0

(-2)x^2+(5)x+(-15)=0

x=(-5+-sqrt(5^2-(4*-2*-15)))/(2*-2)

x=(-5+-sqrt(-95))/-4

x=(-5+-isqrt95)/-4