problem: For what real value of $k$ is $\frac{13-\sqrt{131}}{4}$ a root of $2x^2-13x+k$?
solution: We could substitute $(13-\sqrt{131})/4$ for $x$ in the equation, but the quadratic formula suggests a quicker approach. Substituting $2$, $-13$, and $k$ into the quadratic formula gives  \[
  \frac{-(-13)\pm\sqrt{(-13)^2-4(2)(k)}}{2(2)}= \frac{13\pm\sqrt{169-8k}}{4}.
  \]Setting $(13+\sqrt{169-8k})/4$ and $(13-\sqrt{169-8k})/4$ equal to $(13-\sqrt{131})/4$, we find no solution in the first case and $169-8k=131$ in the second case.  Solving yields $k=(169-131)/8=38/8=\boxed{\frac{19}{4}}$.