Solving $C^{12}_{x} + C^{12}_{x+1} = C^{13}_{2x}$. $C^{12}_{x} + C^{12}_{x+1} = C^{13}_{2x}$
I did find by brute force the solutions $n=1$ and $n=4$, through the inequalities $2x \le 13, x \ge 0 \implies x \in \{0,1,2,3,4,5,6\}$
But is there a more analytical way to solve this? 
Here is my attempt:
$C^{12}_{x} + C^{12}_{x+1} = C^{13}_{2x}$
$C^{13}_{x+1} = C^{13}_{2x}$
$\dfrac{13!}{(x+1)!(13-x-1)!}=\dfrac{13!}{(2x)!(13-2x)!}$
$\dfrac{(2x)!}{(x+1)!}=\dfrac{(13-x-1)!}{(13-2x)!}$
$\dfrac{(2x)(2x-1)...(2x-x)(2x-(x-1))!}{(x+1)!}=\dfrac{(13-x-1)!}{(13-2x)!}$
$(2x)(2x-1)...x=\dfrac{(13-x-1)(13-x-2)...(13-x-(x+1))(13-2x)!}{(13-2x)!}$
$(2x)(2x-1)...x=(13-x-1)(13-x-2)...(13-2x+1)$
Wolfram says these products can be written in terms of Pochhammer symbols as:
$x(x+1)_{x} = \dfrac{2(x-6)(13-2x)_{x+1}}{x-13}$
But I have no idea how to solve this.
Also, is by numerical methods the only way to get the non-natural solutions $-9,-8,-7,-6,-5,-4,-3,-2$ ?
Thanks.

Using the property 

$$C_r^n+C_{r-1}^n=C_r^{n+1}$$

On the LHS, with $n=12,r=x+1$ you have
$$C^{13}_{x+1}=C^{13}_{2x}$$
You now have either $x+1=2x$ or $2x+(x+1)=13$.
From here, it's fairly easy to get $x=1,4$.
