Solve (c)(2xy+1)^2-x^2+2=0

$$( c ) ( 2 x y + 1 ) ^ { 2 } - x ^ { 2 } + 2 = 0$$

Solve for c (complex solution)

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

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2xy+1\right)^{2}.

$$c\left(4x^{2}y^{2}+4xy+1\right)-x^{2}+2=0$$

Use the distributive property to multiply c by 4x^{2}y^{2}+4xy+1.

$$4cx^{2}y^{2}+4cxy+c-x^{2}+2=0$$

Add x^{2} to both sides. Anything plus zero gives itself.

$$4cx^{2}y^{2}+4cxy+c+2=x^{2}$$

Subtract 2 from both sides.

$$4cx^{2}y^{2}+4cxy+c=x^{2}-2$$

Combine all terms containing c.

$$\left(4x^{2}y^{2}+4xy+1\right)c=x^{2}-2$$

Divide both sides by 4x^{2}y^{2}+4xy+1.

$$\frac{\left(4x^{2}y^{2}+4xy+1\right)c}{4x^{2}y^{2}+4xy+1}=\frac{x^{2}-2}{4x^{2}y^{2}+4xy+1}$$

Dividing by 4x^{2}y^{2}+4xy+1 undoes the multiplication by 4x^{2}y^{2}+4xy+1.

$$c=\frac{x^{2}-2}{4x^{2}y^{2}+4xy+1}$$

Divide x^{2}-2 by 4x^{2}y^{2}+4xy+1.

$$c=\frac{x^{2}-2}{\left(2xy+1\right)^{2}}$$

Solve for c

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

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2xy+1\right)^{2}.

$$c\left(4x^{2}y^{2}+4xy+1\right)-x^{2}+2=0$$

Use the distributive property to multiply c by 4x^{2}y^{2}+4xy+1.

$$4cx^{2}y^{2}+4cxy+c-x^{2}+2=0$$

Add x^{2} to both sides. Anything plus zero gives itself.

$$4cx^{2}y^{2}+4cxy+c+2=x^{2}$$

Subtract 2 from both sides.

$$4cx^{2}y^{2}+4cxy+c=x^{2}-2$$

Combine all terms containing c.

$$\left(4x^{2}y^{2}+4xy+1\right)c=x^{2}-2$$

Divide both sides by 4x^{2}y^{2}+4xy+1.

$$\frac{\left(4x^{2}y^{2}+4xy+1\right)c}{4x^{2}y^{2}+4xy+1}=\frac{x^{2}-2}{4x^{2}y^{2}+4xy+1}$$

Dividing by 4x^{2}y^{2}+4xy+1 undoes the multiplication by 4x^{2}y^{2}+4xy+1.

$$c=\frac{x^{2}-2}{4x^{2}y^{2}+4xy+1}$$

Divide x^{2}-2 by 4x^{2}y^{2}+4xy+1.

$$c=\frac{x^{2}-2}{\left(2xy+1\right)^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{-2cy+\sqrt{2+c-8cy^{2}}}{4cy^{2}-1}\text{; }x=-\frac{2cy+\sqrt{2+c-8cy^{2}}}{4cy^{2}-1}\text{, }&y=0\text{ or }c\neq \frac{1}{4y^{2}}\\x=-\frac{c+2}{4cy}\text{, }&c=\frac{1}{4y^{2}}\text{ and }y\neq 0\end{matrix}\right.$$