Question 1164700
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<pre>

n sides and  {{{(n*(n-3))/2}}}  diagonals.


This quantities are equal (given)


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


Simplify


    2n = n*(n-3)

    n^2 - 3n - 2n = 0

    n^2 - 5n = 0

    n*(n-5) = 0.


The roots are  n= 0  and  n= 5.


Of them, only n= 5 is interesting for us.


It is the problem's answer.


<U>ANSWER</U>.  5-sided polygon (pentagon).
</pre>

Solved.