This is basically asking you to find the number of vertices \(n\) of a complete graph that has 91 edges. (I've attached an example of one where \(n=7\), taken from Wiki "complete graph" article)
Let's say you have \(n\) unconnected vertices, each labelled with a number \(v_1,v_2,\ldots,v_n\) (just for the purpose of tracking which ones you've already taken into account). Draw edges connecting \(v_1\) to all the other vertices. There are \(n-1\) possible edges you can draw.
Now starting from \(v_2\), there's already an edge between it and \(v_1\), which means there are \(n-2\) other possible edges that you can draw.
Next, from \(v_3\), you can only draw \(n-3\) new edges because \(v_3\) is already connected to \(v_1\) and \(v_2\).
Continuing in this pattern, you find that the second-to-last edge, \(v_{n-1}\) can only be connected once to \(v_n\), and finally arriving at \(v_n\) you find that all possible connections have been exhausted.
The takeaway here is that the total number of possible connections is the sum
\(S=(n-1)+(n-2)+(n-3)+\cdots+1+0\)
Writing the terms in reverse order, you have
\(S=0+1+2+\cdots+(n-2)+(n-1)\)
So now if you add up the first, second, ..., terms of the two equivalent sums, you have
\(2S=(n-1+0)+(n-2+1)+(n-3+2)+\cdots+(1+n-2)+(0+n-1)\)
\(2S=(n-1)+(n-1)+(n-1)+\cdots+(n-1)+(n-1)\)
In other words, you're adding up \(n-1\), \(n\) times (since each \(n-1\) corresponds to a distinct vertex), so you have
\(2S=n(n-1)\)
Dividing by 2, you end up with
\(S=\dfrac{n(n-1)}2\)
Now, since you know that 91 handshakes took place in total, you have
\(91=\dfrac{n(n-1)}2\implies 182=n(n-1)\implies n=-14,13\)
Obviously, we omit the negative solution, so there were 13 people at the meeting.
