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

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} > 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 < 0 $ and the other side  of the inequality is surely positive. 
