find the solution set of the following inequality with so many radical I get lost
$\sqrt[4]{\frac{\sqrt{x^{2}-3x-4}}{\sqrt{21}-\sqrt{x^{2}-4}}}\geqslant x-5$
Edit
I get online with wolfram
 -5 < x <= -2 || x == -1 || 4 <= x < 5

Well, this is a very messy problem.
We start off with:
$\sqrt[4]{\frac{\sqrt{x^{2}-3x-4}}{\sqrt{21}-\sqrt{x^{2}-4}}}\geqslant x-5$
Now we can raise each side to the power $4$. 
${\frac{\sqrt{x^{2}-3x-4}}{\sqrt{21}-\sqrt{x^{2}-4}}}\geqslant (x-5)^4$
We can then multiply the denominator to the right side:
$\sqrt{x^{2}-3x-4}\geqslant(x-5)^4 * (\sqrt{21}-\sqrt{x^{2}-4})$
From here, again we can square both sides, receiving:
${x^{2}-3x-4}\geqslant(x-5)^8 * (\sqrt{21}-\sqrt{x^{2}-4})²$
Afterwards, you can expand these.
