I can't read that language but I'll guess it says write a second degree equation then solve for n when d=10.
[tex]10 = \dfrac{n(n-3)}{2}[/tex]
[tex]20 = n^2 - 3n[/tex]
[tex]n^2 - 3n - 20 = 0[/tex]
Answer: n² - 3n - 20 = 0
That doesn't factor so there is no integer n solution.
That means there are no polygons with 10 diagonals.
[tex]n = \frac 1 2(3 \pm \sqrt{89})[/tex]
