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!

Edit
First see the comment of NinadMunshi immediately following this answer.  I have edited the answer accordingly.

Continuing Infinity_hunter's answer, and letting
$$x_n~ \text{denote}~ 
\sqrt{12 + \sqrt{12 +  \cdots  \text{n times} }}
$$
To show convergence, all that is necessary is to show that

*

*the sequence is bounded

*the sequence is strictly increasing.

Clearly, $0 < x_1 < 4.$  Assume that $0 < x_n < 4.$  Then
$$x_{(n+1)} = \sqrt{12 + x_n} < \sqrt{12 + 4} < 4$$
and
$$x_{(n+1)} = \sqrt{12 + x_n} > \sqrt{12 + 0} > 0.$$
Therefore,
$$0 < x_n < 4 \implies 0 < x_{(n+1)} < 4.$$
To show that the sequence is (therefore) strictly increasing:
Since
$$ 0 < x_n < 4$$
and
$$[(x_n)^2 - x_n - 12] = (x_n - 4)(x_n + 3)$$
I conclude that
$$[(x_n)^2 - x_n - 12] < 0 \implies (x_n)^2 < x_n + 12.$$
However, by the definition of the sequence
$$\left[x_{(n+1)}\right]^2 = x_n + 12.$$
Therefore
$$\left[x_{(n+1)}\right]^2 > (x_n)^2.$$
Therefore, since each element in the sequence is positive,
$$x_{(n+1)} > (x_n).$$
Therefore, the sequence is strictly increasing.
