How do you find the zeros, real and imaginary, of #y=-x^2+7x+11# using the quadratic formula?

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Answer 1 · 2017-12-13T12:33:44

#x=(7+sqrt(93))/2 or (7-sqrt(93))/2#

The quadratic formula is:
#x=(-b+-sqrt(b^2-4ac))/(2a)#

Putting our values in gives us:
#x=(-7+-sqrt(7^2-4(-1*11)))/(-2)#

#=(-7+-sqrt(49-4(-11)))/(-2)#

#=(-7+-sqrt(49+44))/(-2)#

#=(-7+-sqrt(93))/-2#

#=(-7-sqrt(93))/-2 or (-7+sqrt(93))/-2#

#=(7+sqrt(93))/2 or (7-sqrt(93))/2#