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