How to find the values that $x$ can take to be a real number? 
$$3 \cdot \sqrt{x+4} + 5 \cdot \sqrt [8]{6-x}$$
  - How to find the values that $x$ can take to be a real number? 

I'm a bit confused. However, I want to show my thinkings:
$$ x + 4 > 0 \implies  x > -4$$
and 
$$6-x>0 \implies  6 >x$$
My Kindest Regards,

You are almost correct! You just haven't considered the case that the radicals could be $0$. 
We have $$\sqrt{x+4}=0\implies x=-4$$ and $$\sqrt[8]{6-x}=0\implies 6=x$$
Hence your inequalities become $$x\ge-4,\,x\le6$$ or more succinctly $$-4\le x\le6$$
