Question 602117
Let d = number of diagonals


The formula to find the number of diagonals 'd' is {{{d=(n(n-3))/2}}}



Solve this for n to find the number of sides given the number of diagonals



{{{d=(n(n-3))/2}}}


{{{2d=n(n-3)}}}


{{{2d=n^2-3n}}}


{{{0=n^2-3n-2d}}}


{{{n^2-3n-2d=0}}}


Now use the quadratic formula to solve for 'n'


{{{n = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{n = (-(-3)+-sqrt((-3)^2-4(1)(-2d)))/(2(1))}}}


{{{n = (3+-sqrt(9+8d))/(2)}}}


{{{n = (3+sqrt(9+8d))/(2)}}} or {{{n = (3-sqrt(9+8d))/(2)}}}


So things are a bit uglier, but now you have a formula for 'n' given the number of diagonals 'd'.