Question

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

Answer 1

`x=-1/3+2/3sqrt(2)i` or `x=-1/3-2/3sqrt(2)i`

Explanation:

Given
`3x^2-2x+3=0`
dividing by `3`
`x^2-2/3x+1=0`
Using the quadratic formula
`x_{1,2}=-1/3pmsqrt(1/9-9/9)`
this is
`x_{1,2}=-1/3pm\sqrt(8/9)i`
simplifying
`x_{1,2}=-1/3\pm2/3\sqrt(2)i`