Question

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

Answer 1

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

Explanation:

`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`