# Polygon 42

Which polygon has 42 more diagonals than sides?

Result

n =  12

#### Solution:

$\dfrac12n (n-3) = n+42 \ \\ n^2 -5n -84 =0 \ \\ \ \\ a=1; b=-5; c=-84 \ \\ D = b^2 - 4ac = 5^2 - 4\cdot 1 \cdot (-84) = 361 \ \\ D>0 \ \\ \ \\ n_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ 5 \pm \sqrt{ 361 } }{ 2 } \ \\ n_{1,2} = \dfrac{ 5 \pm 19 }{ 2 } \ \\ n_{1,2} = 2.5 \pm 9.5 \ \\ n_{1} = 12 \ \\ n_{2} = -7 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (n -12) (n +7) = 0 \ \\ \ \\ n>0 \ \\ n = 12$

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