problem: Find the value of $\sqrt{12-\!\sqrt{12-\!\sqrt{12-\cdots}}}$.
solution: 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 $x = \boxed{3}$.