Solve 7x^2-6y*2x+3=20z^4

$$7 { x }^{ 2 } -6y \times 2x+3=20 { z }^{ 4 }$$

Solve for y

$$\left\{\begin{matrix}y=\frac{7x^{2}-20z^{4}+3}{12x}\text{, }&x\neq 0\\y\in \mathrm{R}\text{, }&x=0\text{ and }\frac{\sqrt[4]{1500}}{10}\geq \frac{20^{\frac{3}{4}}\sqrt[4]{3}}{20}\text{ and }|z|=\frac{\sqrt[4]{1500}}{10}\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply 6 and 2 to get 12.

$$7x^{2}-12yx+3=20z^{4}$$

Subtract 7x^{2} from both sides.

$$-12yx+3=20z^{4}-7x^{2}$$

Subtract 3 from both sides.

$$-12yx=20z^{4}-7x^{2}-3$$

The equation is in standard form.

$$\left(-12x\right)y=20z^{4}-7x^{2}-3$$

Divide both sides by -12x.

$$\frac{\left(-12x\right)y}{-12x}=\frac{20z^{4}-7x^{2}-3}{-12x}$$

Dividing by -12x undoes the multiplication by -12x.

$$y=\frac{20z^{4}-7x^{2}-3}{-12x}$$

Divide 20z^{4}-7x^{2}-3 by -12x.

$$y=\frac{-\frac{5z^{4}}{3}+\frac{1}{4}}{x}+\frac{7x}{12}$$

Solve for x

$$x=\frac{\sqrt{36y^{2}+140z^{4}-21}+6y}{7}$$

$$x=\frac{-\sqrt{36y^{2}+140z^{4}-21}+6y}{7}\text{, }|z|>\frac{20^{\frac{3}{4}}\sqrt[4]{3}}{20}\text{ or }|y|\geq \frac{\sqrt{21-140z^{4}}}{6}$$