How do you find the roots, real and imaginary, of #y=-2(x +3)^2-5x + 6# using the quadratic formula?

1 Answer
Feb 16, 2016

Answer:

Roots: #x in {-7.723, -0.777}#

Explanation:

First convert the given equation:
#y=-2(x+3)^2-5x+6# into standard quadratic form:

#y=-2(x^2+6x+9)-5x+6#

#y=-2x^2-12x-18-5x+6#

#y=-2x^2-17x-12#

Apply the quadratic formula:
#color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)#
with #a=-2#, #b=-17#, and #c=-12#

#x=(17+-sqrt(17^2-4(2)(12)))/(2(-2)#
which (with a bit of arithmetic) gives:
#color(white)("XXX")x~~-7.723# or #x~~-0.777#