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

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Answer 1 · 2018-03-15T05:54:26

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

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#