Solve (2y^2-4x+5)dx(2y-4xy-4)dy=0

$$( 2 y ^ { 2 } - 4 x + 5 ) d x ( 2 y - 4 x y - 4 ) d y = 0$$

Solve for d (complex solution)

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{C}\text{, }&x=0\text{ or }y=0\text{ or }\left(y=\frac{2}{1-2x}\text{ and }x\neq \frac{1}{2}\right)\text{ or }x=\frac{y^{2}}{2}+\frac{5}{4}\end{matrix}\right.$$

Solve for x (complex solution)

$$\left\{\begin{matrix}\\x=0\text{; }x=\frac{y^{2}}{2}+\frac{5}{4}\text{, }&\text{unconditionally}\\x=\frac{1}{2}-\frac{1}{y}\text{, }&y\neq 0\\x\in \mathrm{C}\text{, }&d=0\text{ or }y=0\end{matrix}\right.$$

Solve for d

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{R}\text{, }&x=0\text{ or }y=0\text{ or }\left(y=\frac{2}{1-2x}\text{ and }x\neq \frac{1}{2}\right)\text{ or }x=\frac{y^{2}}{2}+\frac{5}{4}\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}\\x=0\text{; }x=\frac{y^{2}}{2}+\frac{5}{4}\text{, }&\text{unconditionally}\\x=\frac{1}{2}-\frac{1}{y}\text{, }&y\neq 0\\x\in \mathrm{R}\text{, }&d=0\text{ or }y=0\end{matrix}\right.$$