Question

$$2x^{2}+2y^{2}\cdot16x+4y+20=0$$

Solve for x (complex solution)

$x=\sqrt{64y^{4}-2y-10}-8y^{2}$
$x=-\sqrt{64y^{4}-2y-10}-8y^{2}$

Solve for y (complex solution)

$\left\{\begin{matrix}y=\frac{\sqrt{1-160x-16x^{3}}-1}{16x}\text{; }y=-\frac{\sqrt{1-160x-16x^{3}}+1}{16x}\text{, }&x\neq 0\\y=-5\text{, }&x=0\end{matrix}\right.$

Solve for x

$x=\sqrt{64y^{4}-2y-10}-8y^{2}$
$x=-\sqrt{64y^{4}-2y-10}-8y^{2}\text{, }1024y^{4}-32y-160\geq 0$

Solve for y

$\left\{\begin{matrix}y=\frac{\sqrt{1-160x-16x^{3}}-1}{16x}\text{; }y=-\frac{\sqrt{1-160x-16x^{3}}+1}{16x}\text{, }&x\neq 0\text{ and }1-160x-16x^{3}\geq 0\\y=-5\text{, }&x=0\end{matrix}\right.$