Question

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

Answer 1

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

Explanation:

`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`