Diagonals

What x-gon has 54 diagonals?

Correct result:

x =  12

Solution:

$$\begin{gathered} 54=n(n-3)\mathrm{/}2 \\ -0.5n^2+1.5n+54=0 \\ 0.5n^2-1.5n-54=0 \\ a=0.5;b=-1.5;c=-54 \\ D=b^2-4ac=1.5^2-4\cdot 0.5\cdot (-54)=110.25 \\ D>0 \\ {n}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{1.5\pm \sqrt{110.25}}{1}} \\ {n}_{1,2}=1.5\pm 10.5 \\ {n}_1=12 \\ {n}_2=-9 \\ \text{ Factored form of the equation: } \\ 0.5(n-12)(n+9)=0 \\ x>0 \\ x=n_1=12 \end{gathered}$$

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