Let $y=\sqrt{x+5}$, then $y^2=x+5$.

The given equation becomes $y=x^-5 \implies x^2 = y+5$.

Substracting these two relationship gives $$x^2-y^2 = y-x\implies (y-x)(y+x+1)=0$$

if $y-x=0$, then $$x=\sqrt{x+5}\implies x=\frac{1\pm\sqrt{21}}{2}$$

Because $x=y=\sqrt{x+5}\ge 0$, we must discard $\frac{1-\sqrt{21}}{2}$.

If $y+x+1=0$, then $$\sqrt{x+5} + x + 1 = 0\implies x=\frac{-1\pm\sqrt{17}}{2}$$

Because $x= -y-1 = -\sqrt{x+5}-1\le -1$, we must discard $\frac{-1+\sqrt{17}}{2}$.

Therefore the given equation has two solutions: $$x_1= \frac{1+\sqrt{21}}{2}, x_2 = \frac{-1-\sqrt{17}}{2}$$