Lot of Square roots

Let $2+ \sqrt{2 + \sqrt{2 + \sqrt{2 + ...}}} = x$
Next, subtract $2$ from both sides and square both sides, the equation becomes:
$2+ \sqrt{2 + \sqrt{2 + \sqrt{2 + ...}}} = (x-2)^{2}$ .
Notice how the left side becomes $x$ again, and by substituting on the left side and distributing on the right side, we get the following equation:
$x= x^{2} - 4x +4$
$x^{2} -5x +4 =0$
$(x-4)(x-1)=0$ so $x= {1,4}$ .
Obviously, the answer must be greater than $2$ because the first term is $2$ and we're adding positive values. Knowing that x must be greater than $2$ , we can eliminate $1$ from the solution set and that leaves us with the final answer: $\boxed{4}$ .

Move the 2 over to form $\sqrt{2+\sqrt{2+\sqrt{2+...}}} = -2$ and square both sides $(\sqrt{2+\sqrt{2+\sqrt{2+...}}})^{2} = (-2)^{2}$ . What you will get is $2 + \sqrt{2+\sqrt{2+\sqrt{2+...}}} = 4$ , which is what we want. Thus our answer is $\boxed{4} \textbf{ Ans.}$