How do you find the zeros, real and imaginary, of #y=-x^2 -2x +5# using the quadratic formula?

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Answer 1 · 2017-11-05T11:27:20

The quadratic formula says that the zeros of a quadratic function can be calculated as:

#x_1=(-b-sqrt(Delta))/(2a)#

and

where

Here we have:

#Delta=(-2)^2-4*(-1)*5=4+20=24#

#sqrt(Delta)=sqrt(24)=2sqrt(6)#

The discriminant #(Delta)# is greater than zero, so the function has 2 different real zeros:

#x_1=(2-2sqrt(6))/(-2)=-1+sqrt(6)#

#x_1=(2+2sqrt(6))/(-2)=-1-sqrt(6)#