find n, if the coefficient of $x^2$ in $(3 + 2x)^n$ is 20412. Find $n$, if the coefficient of $x^2$ in $(3+2x)^n$ is 20412
I solved it using binomial theorem and got 
$n(n-1)3^{n-2} = 10206$
The term like $n(n-1)3^{n-2}$, I am not sure how to solve it.
 A: The coefficient should be $2n(n-1)3^{n-2}$.
To obtain $n$ use prime factorisation: $\;20412=2^2\cdot3^6\cdot 7$. Maybe $n$ or $n-1$ is divisible by $3$, so the equality
$$2n(n-1)3^{n-2}=2^2\cdot3^6\cdot 7$$
implies $\;n\le 8$. 
By Euclid's lemma, it also implies $7$ divides $n$ or $n-1$. Combining with $n\le 8$, we have that $n=7$ or $n=8$. However $n=8$ is impossible, as the expression is divisible only by $2^2$. Thus there is only one solution: $\;n=\color{red}7$.
A: The question asks to find the $n$ so that
$$
\binom{n}{2}3^{n-2}(2x)^2=20412x^2\tag{1}
$$
That is
$$
2n(n-1)3^{n-2}=20412\tag{2}
$$
Plugging in values of $n$ leads to a solution pretty quickly.

We can also try to apply the Lambert W Function. Note that $n(n-1)\sim\left(n-\frac12\right)^2$. Therefore, multiplying $(2)$ by $\frac{3^{3/2}}2$, taking the square root, and multiplying by $\frac{\log(3)}2$, we want to solve
$$
\left(n-\frac12\right)\frac{\log(3)}2\,e^{\left(n-\frac12\right)\frac{\log(3)}2}\approx\frac{\log(3)}2\sqrt{20412\cdot\frac{3^{3/2}}2}\tag{3}
$$
That is,
$$
\left(n-\frac12\right)\frac{\log(3)}2
\approx\operatorname{W}\left(\frac{\log(3)}2\sqrt{20412\cdot\frac{3^{3/2}}2}\right)\tag{4}
$$
Computing
$$
\begin{align}
n
&\approx\frac12+\frac2{\log(3)}\operatorname{W}\left(\frac{\log(3)}2\sqrt{20412\cdot\frac{3^{3/2}}2}\right)\\
&=6.9957802\tag{5}
\end{align}
$$
Checking $n=7$ in $(2)$ shows that that is the solution.

Lacking an appropriate CAS, an algorithm is given in this answer to compute Lambert W.
