Find all pairs (x,y) of real numbers such that x + y = 10 and x^2 + y^2 = 56.

Note that $(x+y)^2-2xy=100-2xy=x^2+y^2=56\rightarrow 100-2xy=56\rightarrow xy=22, x+y=10$. So x and y are the roots to this quadratic: $z^2-10z+22=0\rightarrow z^2-10z+25=3\rightarrow (z-5)^2=3\rightarrow z=5\pm \sqrt{3}\rightarrow\boxed{(x,y)=(5+\sqrt3,5-\sqrt3), (5-\sqrt3, 5+\sqrt3)}$.