I am trying to solve the following equation by finding '$\alpha$', but unfortunately I can't go further, I appreciate any help.
$10=\dfrac{\Gamma(1+2/\alpha)}{\Gamma(1+1/\alpha)^2} +1$
Since $\Gamma(1+x)=x!$ and by considering $x=1/\alpha$
$10=\dfrac{(2x)!}{(x!*x!)}$
and then?
and I know that how I can find x if I know y in $\Gamma(x)=y\implies x=\Gamma^{(-1)}(y)$ from What is X if I know y and Gamma(X)=y? , but I can not get to this point.
@Claude, Thank you for your support, I think I did it ( in R by uniroot),
ufun <- function(x,up2) { uniroot(function(z) ((gamma(1+2*z)/gamma(1+z)^2-x)) ,c(0,up2))$root}
irr <- 0
up2=0.009
while(irr<=0)
{
up2=up2+0.001
irr <- NA
irr=try( ufun(3,up2),silent=T)
if(!is.numeric(irr)) irr <- 0
else
print(irr)
}
As said in comments, numerical methods are required.
As you suggested, changing variable $\alpha=\frac 1x$ makes the equation to be $$9=\dfrac{\Gamma(1+2x)}{\Gamma(1+x)^2}$$ It would be better conditionned taking logarithms to look for the zero of $$f(x)=\log\left(\Gamma(1+2x) \right)-2\log\left(\Gamma(1+x) \right)-\log(9)$$ By inspection $$f(0)=-\log (9)\qquad f(1)=-\log \left(\frac{9}{2}\right)\qquad f(2)=-\log \left(\frac{3}{2}\right)\qquad f(3)=\log \left(\frac{20}{9}\right)$$ So, the root is between $2$ and $3$.
Now, using Taylor around $x=2$ gives $$f(x)=(\log (6)-\log (9))+\frac{7 (x-2)}{6}+\left(\frac{\pi ^2}{6}-\frac{115}{72}\right) (x-2)^2+O\left((x-2)^3\right)$$ Solving the quadratic leads to $x\approx 2.34274$.
Using Taylor around $x=3$ gives $$f(x)=(\log (20)-\log (9))+\frac{37 (x-3)}{30}+\left(\frac{\pi ^2}{6}-\frac{973}{600}\right) (x-3)^2+O\left((x-3)^3\right)$$ Solving the quadratic leads to $x\approx 2.34445$.
It is probably sufficient (the "exact" solution being $x\approx 2.34315$).
Edit
If you are very lazy, using the values for $x=1,2,3$ to which correspond $y=2,6,20$ write the equation of a parabola. This should give $$y=5x^2-11x+8$$; so, if you want $y=9$, you need to solve $5x^2-11x-1=0$ the positive solution being $x=\frac{11+\sqrt{141}}{10} \approx 2.28743$ which gives a "reasonable" approximation.
Update
If you need to solve accurately for $x$ the equation $$a=\dfrac{\Gamma(1+2x)}{\Gamma(1+x)^2}$$ numerical methods are required and Newton would be the simplest, assuming a "reasonable" starting point.
However, since the function is varying very fast, I would again suggest to search for the zero of $$f(x)=\log\left(\Gamma(1+2x) \right)-2\log\left(\Gamma(1+x) \right)-\log(a)$$ $$f'(x)=2\left( \psi (2 x+1)-\psi (x+1)\right)$$ $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Concerning the starting point, for "small" values of $x$, we could use the Padé approximant built around $x=0$ to get $$\log\left(\Gamma(1+2x) \right)-2\log\left(\Gamma(1+x) \right)\approx \frac{\pi ^4 x^2}{72 \zeta (3)x+6 \pi ^2}$$ which, for $a=9$ would give $x_0=2.48892$ and the converge would be quite fast as shown below $$\left( \begin{array}{cc} n & x_n \\ 0 & 2.4889169127815621492 \\ 1 & 2.3437475740831992170 \\ 2 & 2.3431531513648521187 \\ 3 & 2.3431531406628368140 \\ 4 & 2.3431531406628368105 \end{array} \right)$$
This approximation is quite good for $0\leq x \leq 2$ (that is to say for $1 \leq a \leq 6$).
For larger values, derived from a linear regression for $2\leq x \leq 13$ (that is to say for $6 \leq a \leq 10^{7}$), we could use $$\log\left(\Gamma(1+2x) \right)-2\log\left(\Gamma(1+x) \right)\approx 1.31146 x-0.973398$$ which, for $a=9$ would give $x_0=2.41763$.
All of that would make the solution reached very fast even for high accuracy.
Let us try it for $a=123456$; the iterates would then be $$\left( \begin{array}{cc} n & x_n \\ 0 & 9.6730500000000000000 \\ 1 & 9.6984246472567273657 \\ 2 & 9.6984234280990188021 \\ 3 & 9.6984234280990159973 \end{array} \right)$$