Formula for the number of diagonals in a regular polygon:
D = (n(n-3)/2)
Here D = number of diagonals which is 20
n = number of sides which is not known
20=(n^2-3n)/2
40=n^2-3n
n^2-3n-40= 0
By factorizing the equation you will get
n^2-8n+5n-40=0
n(n-8)+5(n-8)=0
(n-8)(n+5)= 0
So either
n-8=0 or n+5=0
n=8 or n=-5
n can not be negative therefore
n = 8 sides or regular Octagon
D = (n(n-3)/2)
Here D = number of diagonals which is 20
n = number of sides which is not known
20=(n^2-3n)/2
40=n^2-3n
n^2-3n-40= 0
By factorizing the equation you will get
n^2-8n+5n-40=0
n(n-8)+5(n-8)=0
(n-8)(n+5)= 0
So either
n-8=0 or n+5=0
n=8 or n=-5
n can not be negative therefore
n = 8 sides or regular Octagon
