My, what large powers you have!

$(2x+3)^{20} = ....+[20Cn] (2x)^{n} 3^{20-n} + [20C(n+1)] (2x)^{n+1} 3^{19-n}+...$

$[20Cn] 2^n 3^{20-n} = 2 \times [20C(n+1)] 2^{n+1} 3^{19-n}$

$\frac{20*19*....*(20-(n-1))}{n!} \frac{2^n}{2 * 2^{n+1}} (\frac{3^{20-n}}{3^{19-n}}) = \frac{20*19*.....(20-n)}{(n+1)!}$

$\frac{20*19*....*(20-(n-1))}{n!} \frac{3}{2*2}= \times \frac{20*19*....*(20-(n-1))(20-n)}{(n+1)*n!}$

$\frac{20-n}{n+1} = \frac{3}{4}$

$3(n+1) = 4(20-n)$

$3n+3 = 80-4n$

$(3+4)n= 80-3$

$7n = 77$

$n =\boxed{11}$