Suppose 3sat is solved in 2^sqrt(n). That's subexponential. An n^2 reduction from sat to 3sat is polynomial. However, an algorithm using this reduction with the discovered solution, would run in (2^sqrt(n))^2 which is just 2^n.
but
2^(sqrt(n)^2) = 2^n
This also means your other comment here is wrong to suggest any error: https://news.ycombinator.com/item?id=50254142