Find the number of sides of …

Consider a polygon with n vertices. Each vertex can be connected to the n-1 other vertices but two of those connections (to the two adjacent vertices) form sides of the polygon, not diagonals. So there are n-3 diagonals emanating from each of the n vertices giving n(n-3) diagonals. This formula overcounts because it counts each diagonal twice: once for each vertex on the diagonal. Divide by two to compensate for the overcounting giving n(n-3)/2 diagonals. 

You can set n(n-3)/2 = 20, expand and solve the quadratic equation by factorization but this is time consuming and inefficient. Way faster is to simply plug in the answer choices for n and see which one satisfies the equation: n(n-3) = 40. Ans: n=8.

20=(n^2-3n)/2 

40=n^2-3n 
n^2-3n-40= 0 
Splitting the middle term, 

n^2-8n+5n-40=0 
n(n-8)+5(n-8)=0 
(n-8)(n+5)= 0 

So there are two values for n 
n-8=0 or n+5=0 
n = 8 or n = -5 
n can’t be negative 
So n = 8