jmc

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