jmc

Letting $x=\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots }}}}$, we have $x=\sqrt{12+x}$. Squaring both sides gives $x^2=12+x$, so $x^2-x-12=0$. Factoring the left side gives $(x-4)(x+3)=0$. Therefore, $x=4$ or $x=-3$. Clearly $x$ must be positive, so we have $x=\boxed{4}$.