Question:
Find the minimum value of
\[2x^2 + 2xy + 4y + 5y^2 - x\]over all real numbers $x$ and $y.$

Answer:
We can write the expression as
\begin{align*}
2x^2 + 2xy + 4y + 5y^2 - x &= (x^2 + 2xy + y^2) + \left( x^2 - x + \frac{1}{4} \right) + (4y^2 + 4y + 1) - \frac{1}{4} - 1 \\
&= (x + y)^2 + \left( x - \frac{1}{2} \right)^2 + (2y + 1)^2 - \frac{5}{4}.
\end{align*}We see that the minimum value is $\boxed{-\frac{5}{4}},$ which occurs at $x = \frac{1}{2}$ and $y = -\frac{1}{2}.$