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 · 2016-03-08T09:43:16

Zeros are #7/2-sqrt5/2# and #7/2+sqrt5/2#

Zeros, real and imaginary, of #y=ax^2+bx+c# are given by the quadratic formula #(-b+-sqrt(b^2-4ac))/(2a)#

In #y=−x^2+7x−11#, #a=-1#, #b=7# and #c=-11#

Hence zeros are given by

#(-7+-sqrt(7^2-4xx(-1)xx(-11)))/(2xx(-1))# or

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

#(-7+-sqrt5)/(-2)# or

#7/2+-sqrt5/2#

Hence zeros are #7/2-sqrt5/2# and #7/2+sqrt5/2#