Question 864327
Integers, 2n and 2n+2.
{{{(2n+2)^2-3*2n}}}, the square of the second decreased by three times the first;
{{{(2n+2)^2-6n=46}}}, ... is 46.


{{{4n^2+8n+4-6n=46}}}
{{{4n^2+2n+4-46=0}}}
{{{4n^2+2n-42=0}}}
{{{2n^2+n-21=0}}}
-
{{{(2n+7)(n-3)=0}}}
-
n=3 making the two integers 6 and 8.