SOLUTION: If a polygon of n sides has (n/2)(n - 3) diagonals, how many sides will a polygon with 65 diagonals have? Is there a polygon with 80 diagonals?

(a) We should determine the number n of sides of a polygon
    from equation

        = 65.    (1)


     To solve, multiply both sides by 2 and simplify

        n*(n-3) = 2*65.

        n^2 - 3n - 130 = 0.


     Factor left side

        (n-13)*(n+10) = 0.


     The roots are -10 and 13.  We are looking for positive integer solution, 
     so n= 13 ideally suits.


     ANSWER.  A polygon with 65 diagonals has 13 sides.
.


(b)  We should check if the equation

          = 80    (2)

     has a solution in integer positive numbers n >= 3.


     Again, simplify to that as it was done above in part (a).

         n*(n-3) = 2*80

         n^2 - 3n - 160 = 0.

     The discriminant is  d = b^2 - 4ac = (-3)^2 - 4*1*(-160) = 9 + 640 = 649.


     The number 649 is not a perfect square:   = 25.4754784 . . . is an irrational number.


     Thus, equation (2) has only irrational roots and does not have integer solutions.


     From it, we conclude that a polygon with 80 diagonals does not exist.