How do you find the zeros, real and imaginary, of #y=-x^2 -2x +8# using the quadratic formula?

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Answer 1 · 2016-02-14T11:03:27

#x_1 = -4#
#x_2 = 2#

Given a quadratic equation:
#y = ax^2 + bx + c#

We can find its zeros using the Quadratic Formula:
#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

#y = -x^2 - 2x +8#

#=> x = (-(-2) +- sqrt((-2)^2 - 4*(-1)*(8)))/(2(-1))#

#=> x = (2 +- sqrt(4 - (-32)))/-2#

#=> x = (2 +- sqrt 36)/-2#

#=> x = (2 +- 6)/-2#

#=> x_1 = 8/-2 = -4#

#=> x_2 = -4/-2 = 2#