Find the value of $n$ from the given equation:
$$\binom{n}{n}+2\binom{n}{n-1}+\binom{n}{n-2} = \binom{n+2}{2n-3}$$
I have solved it to get the following and cannot proceed further:
$$\binom{n+2}{2} = \binom{n+2}{2n-3}\\ \frac{(n+2)!}{2!\cdot n!} = \frac{(n+2)!}{(2n-3)!(5-n)!}\\ (2n-3)!(5-n)! = 2!\cdot n!$$
There is another rule (I forgot its name) which says that (for $n, a, b \in \Bbb N$) if:
$$\binom{n}{a} = \binom{n}{b}$$
then either $a = b$ or $n-a = b$.
Apply this rule to your answer $\binom{n+2}{2} = \binom{n+2}{2n-3}$ to get either of the following equations:
$2n-3 = 2$ or
$2n-3 = n$
We reject first one because it gives fractional value for $n$. Only the second one gives valid answer which is $n = 3$.
Of course a smart way (as the smart Michael already posted) is to note that $n+2 \ge 2n-3$ giving $n \le 5$. Checking is very easy then.