Evaluate $\sqrt{12 +\!\sqrt{12 + \!\sqrt{12 + \!\sqrt{12 + \cdots}}}}$.
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 $x= \boxed{4}$.