Hi ;-)
[tex]\text{x}=-1 \ \text{and} \ \text{y}=-4\\\\\dfrac{5x^2-y^2+3}{2x-4y}=\dfrac{5\cdot(-1)^2-(-4)^2+3}{2\cdot(-1)-4\cdot(-4)}=\dfrac{5\cdot1-16+3}{-2+16}=\boxed{-\frac{8}{14}}[/tex]
