Holy Handshakes!

A literal implementation using $Mathematica$

Select[{25,30,12,14},Length@Subsets[Range@#,{2}]==66&]

returns $12$

This situation can be expressed as $66=\frac{n(n-1)}{2}=\sum_{k=1}^{n-1}k$ where $n$ is the number of people present. $66=\frac{n(n-1)}{2}$ $132=n(n-1)$ $12=\frac{n(n-1)}{11}$ $12*11=n(n-1)$ Because $12-1=11$ and 11 is prime and cannot be factored further, we arrive at our solution: $n=12$

The answer is : 12

Here you go,

In general, with n+1 people, the number of handshakes is the sum of the first n consecutive numbers: 1+2+3+ ... + n. Since this sum is n(n+1)/2, we need to solve the equation n(n+1)/2 = 66. This is the quadratic equation n^2+ n -132 = 0. Solving for n, we obtain 11 as the answer and deduce that there were 12 people at the party.