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.
 A: $$9-x^2>x^4+2x^2+1$$
$$x^4+3x^2-8<0$$
$$\left(x^2+\frac32 \right)^2 < 8+\frac94$$
$$\left(x^2+\frac32 \right)^2< \frac{41}4$$
$$0\le x^2<\color{red}-\frac32 + \frac{\sqrt{41}}2$$
Now, your answer should coincide.
A: $\sqrt{9-x^{2}} > x^{2} + 1
\\\textsf{because both the sides are positive, you can square both sides}
\\ 9-x^{2} > (x^2 + 1)^{2}
\\ 9-x^{2} > x^{4} + 2x^{2} + 1
\\ x^{4} + 3x^{2} - 9 < 0
\\ \textsf{assume x^2 = t}
\\ t^2 +3t - 9 < 0
\\ t \in \left(\frac{-3 - \sqrt{41}}{2},\frac{-3 + \sqrt{41}}{2}\right)
\\ x^{2} \in \left(\frac{-3 - \sqrt{41}}{2},\frac{-3 + \sqrt{41}}{2}\right)
\\ x \in \left( -\sqrt{\frac{-3+\sqrt{41}}{2}}, +\sqrt{\frac{-3+\sqrt{41}}{2}} \right)$
A: Hint:

*

*If $A>B>0$ then $A^2>B^2$. Apply this to $\sqrt{9-x^2}>x^2+1$.

*As $x^2+1$ is always positive then $9-x^2>0$.

Both must be satisfied. So you need $\cap$ to intersect the solution sets from both cases.
The hint is exapanded.
\begin{gather}
9-x^2>x^4+2x^2+1\\
x^4+3x^2-8<0\\
(x^2-\frac{-3+\sqrt{41}}{2})(x^2-\frac{-3-\sqrt{41}}{2})<0
\end{gather}
that must be intersected with
\begin{gather}
9-x^2>0\\
x^2<9
\end{gather}
The solution is $x^2< \frac{-3+\sqrt{41}}{2}$ or $$-\sqrt{\frac{-3+\sqrt{41}}{2}}<x< \sqrt{\frac{-3+\sqrt{41}}{2}}$$
