Evaluate \(\sqrt{12 +\!\sqrt{12 + \!\sqrt{12 + \!\sqrt{12 + \cdots}}}}\).
let that expression equal x, then
\(\sqrt{12+x}=x \\ 12 + x = x^2 \\ x^2 -x -12 = 0 \\ (x-4)(x+3) = 0 \\ x = 4,~-3 \\ \text{however as x is a square root x > 0, thus }x=4\)
+1242 
