How do you find the zeros, real and imaginary, of $y=--x^2 +6x +8$ using the quadratic formula?

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Answer 1 · 2018-06-09T09:04:07

$color(purple)(x = (3 - sqrt78/2), 3 + sqrt78/2$

$y = - x^2 + 6x = 8$

$a = -1, b = 6, c = 8$

$x = (-b +- sqrt(b^2 - 4ac)) / (2a)$

$x =( -6 +- sqrt(6^2 + 32)) / -2$

$color(purple)(x = (3 - sqrt78/2), 3 + sqrt78/2$

How do you find the zeros, real and imaginary, of y=--x^2 +6x +8y=--x^2 +6x +8 using the quadratic formula?