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 &lt; x &lt;= -2 || x == -1 || 4 &lt;= x &lt; 5
First let us check where the above formula is defined: 

$ x^2-3x-4 \geq 0  \; \; \Leftrightarrow  \; \; x\in ] -\infty , -1] \cup [4,+\infty[$.

$x^2-4 \geq  0  \; \; \Leftrightarrow  \; \; x\in ] -\infty , -2] \cup [2,+\infty[ $.

$ \sqrt{21 } - \sqrt{ x^2-4} &gt; 0  \; \; \Leftrightarrow  \; \;  x\in   ]-5,-2[ \cup ]2,5[.$ 

Thus  You set of solution must  be subset of  $]-5, -2] \cup [4,5[ $. Note that whenever $x  \in   ]-5, -2] \cup [4,5[ $ the inequality holds true as  $x-5 &lt; 0 $ and the other side  of the inequality is surely positive.
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.