Comment on $x$, if $x=\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$.

$$x=\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}$$$$x^2=6+\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}=6+x$$$$x^2-x-6=(x+2)(x-3)=0$$$$x=-2~or~3$$ $x=-2$ is obviously a bad answer

Therefore, $x=3$

To sate the appetite for rigor of the commenters below: Yes, the series does converge. This is because $3<\sqrt{6+y}<y$ if $y>3$ and $y<\sqrt{6+y}<3$ if $-6<y<3$.

Let $a_{n+1}=\sqrt{6+a_n}$, $a_1=\sqrt{6}$.

We'll prove that $a_n<3$.

Since $a_1<3$, it remains to prove that $a_{n+1}<3$ for $a_n<3$.

Indeed, $a_{n+1}=\sqrt{a_n+6}<\sqrt{3+6}=3$.

Id est, $a_n<3$.

By another hand, $a_{n+1}-a_n=\sqrt{a_n+6}-a_n=\frac{(3-a_n)(2+a_n)}{\sqrt{a_n+6}+a_n}>0$,

which says that there is $\lim\limits_{n\rightarrow+\infty}a_n$.

Let $\lim\limits_{n\rightarrow+\infty}a_n=A$.

Hence, $\lim\limits_{n\rightarrow+\infty}a_{n+1}^2=\lim\limits_{n\rightarrow+\infty}a_n+6$ or $A^2=A+6$ or $A=3$, which gives the answer.

Solution C is correct.

Since $x=\sqrt{6+\sqrt{6+\cdots }}$ we have that:

$$ x=\sqrt{6+x}$$

This reduces to the quadratic equation $x^2-x-6=0$ which has solutions $x=-2$ or $x=3$. $x=-2$ is not a solution since $-2\neq \sqrt{6-2}=\sqrt{4}=2$. Thus $x=3$.