problem: Real numbers $x$ and $y$ have an arithmetic mean of 7 and a geometric mean of $\sqrt{19}$. Find $x^2+y^2$.
solution: The givens tell us that $\frac{x+y}{2}=7$ and $\sqrt{xy}=\sqrt{19}$, or $x+y=14$ and $xy=19$. $(x+y)^2=x^2+2xy+y^2$, so  \[
  x^2+y^2=(x+y)^2-2xy=14^2-2\cdot19=196-38=\boxed{158}
  \]