jmc

We complete the square on the right side by adding $(12/2{)}^2=36$ to both sides: $y+36=x^2+12x+36+5$, so $y+36=(x+6{)}^2+5$, which gives $y=(x+6{)}^2-31$. Since $(x+6{)}^2$ is nonnegative, and $(x+6{)}^2=0$ when $x=-6$, the smallest possible value of $y$ is $\boxed{-31}$.