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!

Method 1 - contraction mapping theorem.
Let $g : [0,\infty) \to [0,\infty)$ be the map $x \mapsto \sqrt{x+12}$. For any $x, y \in [0,\infty)$, we have
$$\begin{align} & g(x) - g(y) = \sqrt{x+12} - \sqrt{y+12} = \frac{x - y}{\sqrt{x+12} + \sqrt{y+12}}\\
\implies & |g(x)-g(y)| \le \frac{|x-y|}{2\sqrt{12}}
\end{align}
$$
This means $g$ is a contraction mapping over $[0,\infty)$.
By Contraction mapping theorem, $g(x)$ have a unique fixed point over $[0,\infty)$. Furthermore, if one pick any $z \in [0,\infty)$ and construct a sequence $z_n$ by $$z_n = \begin{cases}z, &n = 0\\g(z) = \sqrt{z+12}, & n > 0\end{cases}$$
$z_n$ will converges to that fixed point.
Since $g(4) = 4$, that unique fixed point is $4$. By setting $z$ to $0$, we find
$$\begin{array}{rcl}
z_1 &=& \sqrt{12},\\
z_2 &=& \sqrt{12+\sqrt{12}},\\ 
z_3 &=& \sqrt{12+\sqrt{12+\sqrt{12}}}\\
&\vdots&
\end{array}$$
converges to $4$.
Method 2 - explicit bound.
Define $z_n$ as in method $1$ and let $y_n = 4 - z_n$ for $n \ge 1$.
Notice
$$y_{n+1} = 4 - z_{n+1} = 4 - \sqrt{12 + z_n}
= 4 - \sqrt{16-y_n}
= \frac{y_n}{4 + \sqrt{16-y_n}}\tag{*1}
$$
$y_{n+1}$ has same sign as $y_n$. Since $y_1 = 4 - \sqrt{12}> 0$, all $y_n$ are positive.
Notice $\sqrt{16-y_n} = \sqrt{12+z_n} \ge 12$, $(*1)$ implies
$$0 < y_{n+1} < \frac{y_n}{4+\sqrt{12}}$$
Replace $n$ by $1, 2, \ldots, m-1$ and combine the inequalities, we find for $m \ge 1$,
$$0 < y_m \le \frac{y_1}{(4+\sqrt{12})^{m-1}}\quad
\implies\quad 4 - \frac{4}{(4+\sqrt{12})^m}
\le z_m < 4$$
By squeezing,
$$\lim_{m\to\infty} z_m = \lim_{m\to\infty}
\underbrace{\sqrt{12 + \sqrt{12 + \sqrt{12 + \cdots}}}}_{m\text{ times}} = 4$$
