jmc

Letting $x=\sqrt{12-\sqrt{12-\sqrt{12-\cdots }}}$, we have $x=\sqrt{12-x}$. Therefore, $x^2=12-x$, so $x^2+x-12=0$, or $(x+4)(x-3)=0$. Clearly, $x$ must be positive, so $x=\boxed{3}$.