Diagonal formula

<math>D= n(n-3)/2</math>This formula is used to find the amount of diagonals that can be drawn in any polygon. Use n as the sides of the polygon. It also works in reverse by using the amount of diagonals given and factoring.
For example if you have a Heptagon you would do <math>D 7(7-3)/2</math> then you get<math>D7(4)/2 </math> finally you get <math>D28/214</math>.
If you want to use it backwards you can take <math>20 n(n-3)/2</math> then you get <math>40n(n-3)</math> next you use the divison property <math>40=n^2*3n</math> To factor this problem you need to make the equation equal to zero like this <math>n^2*3n-40=0</math> Next you need something that multiplies to get -40, but adds to get -3. In this case it would be <math>(n-8)(n+5)</math> Finally, you treat it as two seperate equations like <math>n-80, n+50</math> it comes out to <math>n8</math> <math>n=-5</math> since a polygon cannot have negative sides the figure would be an octogon.

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