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 $7/2+sqrt5/2$

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