Solve 2x^4+4y^3/2=12

$$2 x ^ { 4 } + 4 y ^ { 3 / 2 } = 12$$

Solve for x (complex solution)

$$x=i\sqrt[4]{6-2y^{\frac{3}{2}}}$$

$$x=\sqrt[4]{6-2y^{\frac{3}{2}}}$$

$$x=-\sqrt[4]{6-2y^{\frac{3}{2}}}$$

$$x=-i\sqrt[4]{6-2y^{\frac{3}{2}}}$$

Solve for y (complex solution)

$$\left\{\begin{matrix}y=\frac{\left(6-x^{4}\right)^{2}}{4}\text{, }&arg(-\frac{x^{4}}{2}+3)<\pi \\y=0\text{, }&x=-\sqrt[4]{6}i\text{ or }x=-\sqrt[4]{6}\text{ or }x=\sqrt[4]{6}i\text{ or }x=\sqrt[4]{6}\end{matrix}\right.$$

Solve for x

$$x=\sqrt[4]{6-2y^{\frac{3}{2}}}$$

$$x=-\sqrt[4]{6-2y^{\frac{3}{2}}}\text{, }y\geq 0\text{ and }y\leq \sqrt[3]{9}$$

Solve for y

$$y=\frac{\sqrt[3]{2}\left(6-x^{4}\right)^{\frac{2}{3}}}{2}$$

$$-\frac{x^{4}}{2}+3\geq 0$$