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 should set
$$ x + 4 \ge 0 \implies  x \ge -4$$
and 
$$6-x\ge0 \implies  6 \le x$$
thus
$$-4\le x\le6$$
