Handshake problem when no. of handshakes are given I've tried to do the following problem:
At the end of a meeting, all participants shake hands with each other. $28$ handshakes were exchanged. How many people attended the meeting?
What I tried:
If there are $n$ people, the first one does $n-1$ handshakes. The second one $n-2$. The third $n-3$ and so on until $1$ handshake. 
So an arithmetic series is there with $n-1$ members, the sum of which is equal to $28$. 
Using
 SUM$=\frac{n-1}{2} \cdot (\text{1st term} + \text{last term}) = 28$
1st term is $n-1$
Last term is $1$
We get $n(n-1)=56$
A quadratic equation is obtained 
$n^2 -n -56 = 0$ 
Edit: $n=8$
I'm sorry I was very sleepy. I couldn't do simple factoring.

By the handshaking-lemma (or by combinatorial argument), the number of handshakes that take place will be $\binom{n}{2} = \frac{n^2-n}{2}$.
If we know the total number of handshakes is 28, this gives us the following equation:
$$28 = \frac{n^2-n}{2}$$
By moving everything to one side of the equation, we have then:
$$0 = n^2-n-56$$
From which we can use the quadratic equation to solve for $n$.

The quadratic equation:  For an equation of the form $ax^2+bx+c=0$, the solutions for $x$ are given as:
$$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

$$n = \frac{1+\sqrt{1+4\cdot 56}}{2} = \frac{1+\sqrt{225}}{2} = \frac{1+15}{2} = 8$$
