Answer: A

Small Hint:

For real square roots the domain requires $x\ge3$

Big Hint:

On that domain, compare $\sqrt{x+4}$ directly with $\sqrt{x-3}$

Solution:

For a real solution, $x\ge3.$ Then $x+4>x-3,$ so $$\sqrt{x+4}>\sqrt{x-3}.$$ Consequently $\sqrt{x+4}-\sqrt{x-3}+1>1,$ and it cannot equal zero. The original equation is a real radical expression, so nonreal roots are outside its domain.

The correct answer is A.