How do you find the zeros, real and imaginary, of $y=-6x^2+3x-9$ using the quadratic formula?

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Answer 1 · 2018-06-09T08:51:14

$color(red)(x = 1/4 - i sqrt207/12, 1/4 + i sqrt207/12$

$y = - 6x^2 + 3x - 9$

$a = -6, b = 3, c = -9$

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

$x = (-3 +- sqrt(9 - 216) ) / -12$

$color(red)(x = 1/4 - i sqrt207/12, 1/4 + i sqrt207/12$

How do you find the zeros, real and imaginary, of y=-6x^2+3x-9y=-6x^2+3x-9 using the quadratic formula?