Question 996978


recall:
Polygon with {{{n}}} sides has {{{n(n-3)/2}}} diagonals

so, the polygon with {{{44}}} diagonals will have:

{{{44 = n(n-3)/2}}}...solve for {{{n}}}

{{{88 = n(n-3)}}}

{{{88 = n^2-3n}}}

{{{n^2-3n -88 = 0}}}...factor completely

{{{n^2-11n+8n -88 = 0}}}

{{{(n^2-11n)+(8n -88) = 0}}}

{{{n(n-11)+8(n -11) = 0}}}

{{{(n-11)(n+8) = 0}}}


solutions:
if {{{(n+8) = 0}}}=>{{{n = -8}}} cannot use this (side length must be positive number)

if {{{(n-11)=0}}} =>{{{n = 11}}}

so, the polygon with {{{44}}} diagonals has {{{11}}} sides