Diagonals

The number of diagonals of a polygon with $n$ sides is $\frac{n(n-3)}{2}$ so setting up the equation and solving we get: $\frac{n(n-3)}{2}=14$ $n(n-3)=28$ Now instead of setting up a quadratic equation and then solving we can simply check factors of $28$ since it is a small number.After checking we get $28=7\times4=7\times(7-3)$ So $n=7$ and $n+1=8$ plugging this value into the formula we get: $\frac{8(8-3)}{2}=\frac{8\times5}{2}=4\times5=\boxed{20}$ For guys who want to see how the quadratic equation gets solved,here it is.Setting up the equation,we get: $n(n-3)=28$ $n^2-3n-28=0$ Plugging the values into the quadratic formula,we get: $\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(-28)}}{2(1)}=\frac{3\pm\sqrt{9+112}}{2}=\frac{3\pm11}{2}$ Solving this ,we get $x=-4\;or\;x=8$ As the number of sides cannot be negative,so $x=8$ plugging this value into the formula we get $20$ the same answer as above

We know the relation between the number of diagonals and sides of a polygon which is given by n(n-3)/2 where n is the number of sides. n(n-3)/2= 14 . By substituting values , we get 2 roots -> -4 and 7. As the sides can't be -4, it is 7. n=7 . n+1=8. 8(8-3)/2= we get THE ANSWER 20.