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The best way-

We have to find $n$ such that $n^2+19n+92=k^2$ where $k$ is any integer. $\Rightarrow n^2+19n+92-k^2=0$ This quadratic equation is satisfied by $n=a, b$ as given by the question.

Thus by Vieta's formula, a+b=-\frac{\text {coefficient of n}}{\text {coefficient of n^2}}=\boxed{-19}

n^2+19n+92 = (n+10)(n+9)+2 =(n+10)^2-n-8 =(n+9)^2+n+9+2 =(n+9)^2+(n+11) So (n+9)^2<=n^2+19n+92<=(n+10)^2 To be a square of an integer we have (n+8)=0 or (n+11)=0 so a= - 8, b= - 11 and a+b= - 19