Solve b=b^2-4x+xc

$$b = b ^ { 2 } - 4 x + x c$$

Solve for c (complex solution)

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

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$b^{2}-4x+xc=b$$

Subtract b^{2} from both sides.

$$-4x+xc=b-b^{2}$$

Add 4x to both sides.

$$xc=b-b^{2}+4x$$

The equation is in standard form.

$$xc=4x-b^{2}+b$$

Divide both sides by x.

$$\frac{xc}{x}=\frac{4x-b^{2}+b}{x}$$

Dividing by x undoes the multiplication by x.

$$c=\frac{4x-b^{2}+b}{x}$$

Solve for c

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

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$b^{2}-4x+xc=b$$

Subtract b^{2} from both sides.

$$-4x+xc=b-b^{2}$$

Add 4x to both sides.

$$xc=b-b^{2}+4x$$

The equation is in standard form.

$$xc=4x-b^{2}+b$$

Divide both sides by x.

$$\frac{xc}{x}=\frac{4x-b^{2}+b}{x}$$

Dividing by x undoes the multiplication by x.

$$c=\frac{4x-b^{2}+b}{x}$$

Solve for b (complex solution)

$$b=\frac{-\sqrt{1+16x-4cx}+1}{2}$$

$$b=\frac{\sqrt{1+16x-4cx}+1}{2}$$

Solve for b

$$b=\frac{-\sqrt{1+16x-4cx}+1}{2}$$

$$b=\frac{\sqrt{1+16x-4cx}+1}{2}\text{, }\left(c\geq 4\text{ or }x\geq -\frac{1}{16-4c}\right)\text{ and }\left(c\leq 4\text{ or }x\leq -\frac{1}{16-4c}\right)$$