How do you find the roots, real and imaginary, of #y=-2(x +1 )^2-x+1# using the quadratic formula?

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Answer 1 · 2018-02-19T21:47:04

The roots are:
#(5+sqrt(17))/(-4), and(5-sqrt(17))/(-4)#

First, expand the equation:

#-2(x^2+2x+1)-x+1#

#-2x^2-4x-2-x+1#

#-2x^2-5x-1#

Now that it is in #ax^2+bx+c# standard form, with

#a=-2,b=-5,c=-1#,

you can plug into the quadratic formula:

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

#(-(-5)+-sqrt((-5)^2-4*-2*-1))/(2*-2)#

#(5+-sqrt(17))/(-4)#

The answer is

#(5+sqrt(17))/(-4), and(5-sqrt(17))/(-4)#