Question

How do you find the zeros, real and imaginary, of `y=9x^2-32+2` using the quadratic formula?

Answer 1

`y=9x^2-32+2=9x^2-30`

To find zeros `9x^2-30=0`

`9x^2=30`

`x^2=30/9=10/3`

`x=+-sqrt(10/3)=+-sqrt10/sqrt3=+-sqrt30/3`

There is no imaginary solutions, both roots are real numbers