Q:

Solving the infinite radical $\sqrt{6+\sqrt{6+2\sqrt{6+3\sqrt{6+...}}}}$

$$\sqrt{6+\sqrt{6+2\sqrt{6+3\sqrt{6+\cdots}}}}$$
This is a modification on the well-known Ramanujan infinite radical, $\sqrt{1+\sqrt{1+2\sqrt{1+3\sqrt{1+\cdots}}}}$, except it cannot be solved by the conventional method -- the functional equation $F(x)^2=ax+(n+a)^2+xF(x+n)$, since setting $n=1$ with $a=0$ requires having $(n+a)^2=1$, not $6$.
Here are some alternative methods I've tried:

The functional equation we have instead for this infinite radical is $F(x)^2=6+xF(x+1)$. I've tried to solve this, but unfortunately it's easy to demonstrate that $F(x)$ cannot be a simple linear function $F(x)=ax+b$. I've tried some slightly more complicated versions -- the equation for a hyperbola, etc. -- but nothing seems to work.
I've tried factoring stuff out from the radical to bring it to a more tenable form. Perhaps not a satisfactorily rigorous approach, I thought of factoring out $\sqrt{6^{N/2}}$ where $N\to\infty$, which allows us to transform the radical into $6^{-N/2}\sqrt{6^{N+1}+\sqrt{6^{2N+1}+2\sqrt{6^{4N+1}+\cdots}}}$, which can be treated as having each term a power of $6^{N/2}$ in the limit. For a radical of the form $\sqrt{\alpha^2+\sqrt{\alpha^4+2\sqrt{\alpha^8+\cdots}}}$ we have the functional equation $F(x)^2=\alpha^{2^x}+xF(x+1)$, or upon letting $F(x)=\alpha^{2^x}p(x)$, you get $p(x)^2-xp(x+1)=\alpha^{-2^x}$, but I'm stuck there.
Similarly, I tried factoring out some arbitrary $N$ then factoring out a term from each radical inside such that the coefficients go from being $1,2,3,\cdots$ to a constant $1/N,1/N,1/N...$, transforming the radical into $N\sqrt{\frac6{N^2}+\frac1N\sqrt{\frac6{N^2}+\frac1N\sqrt{\frac{24}{N^2}+\frac1N\sqrt{\frac{864}{N^2}+\frac1N\sqrt{\frac{1990656}{N^2}+\cdots}}}}}$ where the added terms go as $k_1=6$, $k_{n+1}=\frac{n^2}6k_n^2$. But how might one proceed?
I considered differentiating the function $G(x)=\sqrt{x+\sqrt{x+2\sqrt{x+3\sqrt{x+\cdots}}}}$. But all I got was an equally weird differential equation:

$$\frac{df}{dx}=\frac{1+\frac{1+\frac{1+\frac{{\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}}}{\frac23\frac{\left(\frac{\left(f(x)^2-x\right)^2-x}{2}\right)^2-x}{3}}}{\frac22\frac{\left(f(x)^2-x\right)^2-x}{2}}}{\frac21\left(f(x)^2-x\right)}}{2f(x)}$$
Any ideas as to how I might proceed?/Any alternative (hopefully less tedious, but regardless) methods that might work?

I created a small program to play with this. The exact answer (perhaps as an infinite series) may contain $\sqrt{6}+1/2+...$ somewhere in it, because as you increase the number replacing 6, the radical approaches $\sqrt{x}+1/2$. Of course, this term just comes from the binomial series for $\sqrt{6+\sqrt{6}}$.
I also got nothing on the inverse symbolic calculator.

Here's another possible approach: one may consider the sequence of polynomials:
$$P_1:x^2-6=x$$
$$P_2:\left(\frac{x^2-6}2\right)^2-6=x$$
$$P_3:\left(\frac{\left(\frac{x^2-6}2\right)^2-6}3\right)^2-6=x$$
Formed by taking recurrent approximations to the infinite radical. The limit of $P_n$ as $n\to\infty$ is the root of some function with a power series expansion that can perhaps be calculated in this form. But what is the power series expansion? 
Note that the polynomial gets very complicated very quick. E.g. here's $P_5$:
$$\frac{x^{32}}{2751882854400}-\frac{x^{30}}{28665446400}+\frac{43x^{28}}{28665446400}-\frac{91x^{26}}{2388787200}+\frac{121x^{24}}{191102976}-\frac{53x^{22}}{7372800}+\frac{11167x^{20}}{199065600}-\frac{4817x^{18}}{16588800}+\frac{57659x^{16}}{66355200}-\frac{x^{14}}{1382400}-\frac{9491x^{12}}{1382400}+\frac{367x^{10}}{12800}-\frac{2443x^8}{46080}+\frac{179x^6}{9600}+\frac{2233x^4}{9600}-\frac{71x^2}{160}-x-\frac{33359}{6400}=0$$
See What is the region of convergence of $x_n=\left(\frac{x_{n-1}}{n}\right)^2-a$, where $a$ is a constant?

A:

Currently not correct answer; but I keep it for the record (and hopefully whenever I manage to make any progress on it)
Let $$G:=\sqrt{6+\sqrt{6+2\sqrt{6+3\sqrt{6+\cdots}}}}$$ Then define $$F:=G^2-6=\sqrt{6+2\sqrt{6+3\sqrt{6+\cdots}}}$$ which is easier to work with. Following on from this method, we can immediately match up $n$ and $x$. They are $n=1$ and $x=2$ (as can be observed in the radical).
Finally, we find $a$. The value of $6$ corresponds to $ax+(n+a)^2=2a+(1+a)^2$ so we solve $$6=a^2+4a+1\implies(a-1)(a+5)=0\implies a=1,-5.$$
The result is given as $$F=x+n+a=3+a$$ and since $F$ is clearly non-negative, we have that $a=1$ so $$G=\sqrt{6+F}=\sqrt{6+3+1}=\color{red}{\sqrt{10}}.$$

