Question 1114493
Let {{{N}}} and {{{N+2}}} be the integers.
{{{N(N+2)=10+5(N+N+2)}}}
{{{N^2+2N=10+10N+10}}}
{{{N^2-8N-20=0}}}
{{{(N-10)(N+2)=0}}}
Two solutions:
{{{N-10=0}}}
{{{N=10}}}
10,12
and
{{{N+2=0}}}
{{{N=-2}}}
-2,0