What is the least integer whose square is 48 more than its double?
From the information given, we have $x^2 = 2x + 48$. Rearranging, we get $x^2 - 2x - 48 = 0$, which we can factor as $(x+6)(x-8) = 0$. Therefore, $x = -6\text{ or }8$.  Since we want the lesser, $\boxed{-6}$ is our answer.