Solving $\sqrt{9-x^2} > x^2 + 1$ without graphic calculator for the exact form 
Is there any way to solve this inequality without using a graphic calculator to get the exact form?
$$\sqrt{9-x^2} > x^2 + 1$$

I've tried completing the square but I end up with
$$\frac{3 - \sqrt{41}}{2} < x^2 < \frac{3 + \sqrt{41}}{2}$$
which does not match with the answer on Desmos.

$\sqrt{9-x^{2}} > x^{2} + 1
\\\textsf{because both the sides are positive, you can square both sides}
\\ 9-x^{2} > (x^2 + 1)^{2}
\\ 9-x^{2} > x^{4} + 2x^{2} + 1
\\ x^{4} + 3x^{2} - 9 < 0
\\ \textsf{assume x^2 = t}
\\ t^2 +3t - 9 < 0
\\ t \in \left(\frac{-3 - \sqrt{41}}{2},\frac{-3 + \sqrt{41}}{2}\right)
\\ x^{2} \in \left(\frac{-3 - \sqrt{41}}{2},\frac{-3 + \sqrt{41}}{2}\right)
\\ x \in \left( -\sqrt{\frac{-3+\sqrt{41}}{2}}, +\sqrt{\frac{-3+\sqrt{41}}{2}} \right)$
