Differents ways to evaluate the sum $\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$ 
Evaluate $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$$
My approach:
Let $$x=\sqrt{12+ \sqrt{12+\sqrt{12+\cdots}}}$$
so, we have that $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}\iff \sqrt{12+x}=x \implies 12+x=x^{2} \iff (x+3)(x-4)=0$$
So, the answer is $\boxed{4}$.


Is correct my solution?
Can you show other ways for to solve this problem?
Can you suggest me any  textbooks with similar problems?
Thank you so much!

You need to show the sequence
$$
x_n = \sqrt{12 + \sqrt{12 +  \cdots  \text{n times} }}
$$ converges. Only then you can take $x = \sqrt{12 + \sqrt{12 +\cdots}} $ and continue to other steps.
It is easy to see that $x_n = \sqrt{12 + x_{n-1}}$ with $x_1 = \sqrt {12}$.
As answered by @user2661923 and @QC_QAOA the sequence $x_n$ is bounded. Here is an easy way of showing $x_n$ is strictly increasing. Observe that $x_1 < x_2$ and we assume that $x_{n-1} < x_n$. Then
$$x_{n+1} = \sqrt{12 +x_n} > \sqrt{12 + x_{n-1}} = x_n$$
which completes our inductive argument and we conclude that $x_n$ converges.
Observing that $x_n > 0$ for any $n$ we can conclude that $x \ge0$ by limit theorems. So we must have $x = 4$
