Use algebrad the range of $\displaystyle f(x) = 3x^2+10/x^2-9$

i remember our teacher told us to solve for A, any real number so I did

$\displaystyle 3x^2+10/x^2-9 =A$

$\displaystyle 3x^2+10=A(x^2-9)$

$\displaystyle 3x^2+10=Ax^2-9A$

$\displaystyle 9A+10=Ax^2-3x^2$

$\displaystyle 9A+10=x^2(A-3)$

$\displaystyle 9A+10/A-3 = x^2$

A cannot equal 3

I don't know what the next step is