How do you find the roots, real and imaginary, of #y=-2(x +3)^2-7x+5# using the quadratic formula?

Question

1528 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-30T03:32:59

#x=−8.75780488547035#, #x=−0.7421951145296504#

There are no imaginary solutions.

First, combine like terms and simplify the expression so it's in standard quadratic form:

#ax^2+bx+c#

The original expression:

#-2(x+3)^2-7x+5#

After FOILing #(x+3)#:

#(x+3)(x+3)#

#=x^2+6x+9#

After multiplying that by #-2#:

#-2x^2-12x-18#

After adding additional terms and simplifying:

#-2x^2-19x-13#

To find the roots (AKA zeros), just plug it into the quadratic formula:

#x=(-b\pm\sqrt{b^2-4ac})/(2a)#

#=(9\pm\sqrt{(-19)^2-4(-2)(-13)})/(2(-2)#

#=9\pm\sqrt{257}/(-4)#

#x=−8.75780488547035#, #x=−0.7421951145296504#

There are no imaginary solutions.