Question 113317
Any number that ends in 2 is divisible by 2. 
112=2*56
I don't think that's what you want to do here. 
I think you want to complete the square.
That is, there's a number,d, such that 
{{{(n+d)^2=n^2+2dn+d^2}}}
If you look at your equation and compare the "n" terms, you have
{{{n^2+2n = n^2 +2dn}}}
In your case d=1,
{{{n^2+2n = (n+1)^2-1}}}
And then I can substitute in your original equation,
{{{n^2+2n-112=0}}}
{{{(n+1)^2-1-112=0}}}
{{{(n+1)^2-113=0}}}
{{{(n+1)^2=113}}}
The solutions are 
{{{n+1=113}}}
{{{highlight(n=112)}}}
and
{{{n+1=-113}}}
{{{highlight(n=-114)}}}