Answer
$(-1, -3), (3, 1)$
Work Step by Step
1. Let the two numbers be $x, y$, based on the given conditions, we can get:
$\begin{cases} x-y=2 \\ x^2+y^2=10 \end{cases}$
2. The 1st equation gives $y=x-2$, use it in the 2nd equation to get $x^2+x^2-4x+4=10$ or $x^2-2x-3=0$ or $(x-3)(x+1)=0$, thus $x=-1, 3$
3. Use the 1st equation again to get the solution(s) $(-1, -3), (3, 1)$
