Solve a Quick Math Question: How Many Sides in a Polygon w/ 119 Diagonals?

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The discussion centers on solving for the number of sides in a polygon given that it has 119 diagonals. The formula used is d = n(n-3)/2, where d represents the number of diagonals and n represents the number of sides. By substituting 119 for d, the equation becomes 119 = n(n-3)/2. This leads to the quadratic equation n^2 - 3n - 238 = 0, which can be solved using the Quadratic Formula to find the value of n.

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d=n^2 - 3n / 2

where d = number of diagonals
and n = the number of sides of the polygon.

A polygon has 119 diagonals, how many sides does it have?

119 = n^2 - 3n / 2

I am not sure how to do this? can someone please help?

sorry this isn't physics related
 
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An n-sided polygon has this many diagonals:

[tex]\frac{n(n-3)}{2}[/tex]

Solve for [tex]n[/tex].
 
get_rekd said:
d=n^2 - 3n / 2

where d = number of diagonals
and n = the number of sides of the polygon.

A polygon has 119 diagonals, how many sides does it have?

119 = n^2 - 3n / 2

I am not sure how to do this? can someone please help?

sorry this isn't physics related
Quadratic Formula
 

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