Question

How do you find the zeros, real and imaginary, of `y= 3x^2-6x+14` using the quadratic formula?

Answer 1

`1 +- (isqrt33)/3`

Explanation:

`y = 3x^2 - 6x + 14 = 0`
`D = d^2 = 36 - 168 = - 132 < 0`
Since D < 0, there are no real roots. There are 2 imaginary roots.
`d = +- 2isqrt33`
Two imaginary roots:
`x = -b/(2a) +- d/(2a) = 6/6 +- isqrt33/3 = 1 +- isqrt33/3`