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

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Answer 1 · 2018-05-24T13:23:31

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

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

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