Question

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

Answer 1

`1 +- sqrt3/3`

Explanation:

Use the improved quadratic formula (Google, Yahoo Search)
`D = d^2 = b^2 - 4ac = 36 - 24 = 12 --> d = +- 2sqrt3`
There are 2 real roots:
`x = -b/(2a) +- d/(2a) = 6/6 +- (2sqrt3)/6 = 1 +- sqrt3/3`