Question

In one round robin chess tournament, 36 games were played. How many players entered in the tournament?

Answer 1

$\displaystyle \binom{n}{2}=36\\\\ \dfrac{n!}{2!(n-2)!}=36\\ \dfrac{n(n-1)}{2}=36\\ n(n-1)=72\\ n^2-n-72=0\\ n^2-9n+8n-72=0\\ n(n-9)+8(n-9)=0\\ (n+8)(n-9)=0\\ n=-8 \vee n=9$

The number of players can't be negative, so there was 9 players.