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:
$x=(-7+-sqrt(7^2-4(-1*11)))/(-2)$

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

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

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

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

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

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