Question 1009832
{{{x}}}= number of sides of one of the two polygons
{{{y}}}= number of sides of the other polygons
Since it does not matter which is the one and which is the other, we will find two solutions for {{{x}}} , and the same two solutions for {{{y}}} .
{{{x+y=13}}} , because "the total number of sides of the polygons is 13".
The number of diagonals of a polygon with {{{n}}} sides is {{{(n-3)*n/2}}} ,
so "the sum of the number of diagonals is 25" translates as
{{{(x-3)*x/2+(y-3)*y/2=25}}}<-->{{{(x-3)*x+(y-3)*y=50}}} .
So, we have
{{{system(x+y=13,(x-3)*x+(y-3)*y=50)}}} .
{{{x+y=13}}}-->{{{y=13-x}}}
Substituting into {{{(x-3)*x+(y-3)*y=50}}} , we get
{{{(x-3)*x+(13-x-3)(13-x)=50}}}
{{{x^2-3x+(10-x)(13-x)=50}}} 
{{{x^2-3x+130-10x-13x+x^2=50}}}
{{{2x^2-26x+130=50}}}
{{{2x^2-26x+80=0}}}
{{{x^2-13x+40=0}}}
{{{(x-8)(x-5)=0}}}-->{{{system(x=8,"or",x=5)}}} .
Those are the two solutions.
If {{{x=5}}} , {{{y=13-x=8}}} .
If {{{x=8}}} , {{{y=13-8=5}}} .
One polygon has {{{highlight(5)}}} sides, and the other one has {{{highlight(8)}}} sides.