Question 698952

The formula for the number of diagonals in a polygon with {{{n}}} sides is:

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

Now solve this for {{{d=9}}} diagonals:

{{{n(n -3)/2=9}}}

{{{n(n -3)=9*2}}}

{{{n^2-3n=18}}}

{{{n^2-3n-18=0}}}.....use quadratic formula


{{{n = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


{{{n = (-(-3) +- sqrt( (-3)^2-4*1*(-18) ))/(2*1) }}}


{{{n = (3 +- sqrt( 9+72))/2 }}}


{{{n = (3 +- sqrt(81))/2 }}}

{{{n = (3 +- 9)/2 }}}..we will need only positive solution

{{{n = (3 +9)/2 }}}

{{{n = 12/2 }}}

{{{n = 6}}}


So the number of sides is {{{6}}}.