The number of solutions of the equation $$\sqrt{x+5}+\sqrt{3x+4}=\sqrt{12x+1}$$ is

$$\sqrt{x+5}+\sqrt{3x+4}=\sqrt{12x+1}$$

Squaring once gives

$$x+5+3x+4 +2\sqrt{(x+5)(3x+4)}=12x+1$$

$$\implies \sqrt{(x+5)(3x+4)}=4x-4$$. 

Since the left side is a positive square root, the right side must also be non-negative which means $$x \geq 1$$. Squaring again gives $$13x^2-51x-4=0$$, whose roots are $$x=4$$ and $$x=-\dfrac{1}{13}$$. Only $$x=4$$ satisfies the required condition and the original equation, so there is exactly one solution.