Solve 4b^2=2AC+BC

$$4 b ^ { 2 } = 2 A C + B C$$

Solve for A (complex solution)

$$\left\{\begin{matrix}A=\frac{2b^{2}}{C}-\frac{B}{2}\text{, }&C\neq 0\\A\in \mathrm{C}\text{, }&b=0\text{ and }C=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$2AC+BC=4b^{2}$$

Subtract BC from both sides.

$$2AC=4b^{2}-BC$$

The equation is in standard form.

$$2CA=4b^{2}-BC$$

Divide both sides by 2C.

$$\frac{2CA}{2C}=\frac{4b^{2}-BC}{2C}$$

Dividing by 2C undoes the multiplication by 2C.

$$A=\frac{4b^{2}-BC}{2C}$$

Divide -BC+4b^{2} by 2C.

$$A=\frac{2b^{2}}{C}-\frac{B}{2}$$

Solve for B (complex solution)

$$\left\{\begin{matrix}B=\frac{4b^{2}}{C}-2A\text{, }&C\neq 0\\B\in \mathrm{C}\text{, }&b=0\text{ and }C=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$2AC+BC=4b^{2}$$

Subtract 2AC from both sides.

$$BC=4b^{2}-2AC$$

The equation is in standard form.

$$CB=4b^{2}-2AC$$

Divide both sides by C.

$$\frac{CB}{C}=\frac{4b^{2}-2AC}{C}$$

Dividing by C undoes the multiplication by C.

$$B=\frac{4b^{2}-2AC}{C}$$

Divide -2AC+4b^{2} by C.

$$B=\frac{4b^{2}}{C}-2A$$

Solve for A

$$\left\{\begin{matrix}A=\frac{2b^{2}}{C}-\frac{B}{2}\text{, }&C\neq 0\\A\in \mathrm{R}\text{, }&b=0\text{ and }C=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$2AC+BC=4b^{2}$$

Subtract BC from both sides.

$$2AC=4b^{2}-BC$$

The equation is in standard form.

$$2CA=4b^{2}-BC$$

Divide both sides by 2C.

$$\frac{2CA}{2C}=\frac{4b^{2}-BC}{2C}$$

Dividing by 2C undoes the multiplication by 2C.

$$A=\frac{4b^{2}-BC}{2C}$$

Divide -BC+4b^{2} by 2C.

$$A=\frac{2b^{2}}{C}-\frac{B}{2}$$

Solve for B

$$\left\{\begin{matrix}B=\frac{4b^{2}}{C}-2A\text{, }&C\neq 0\\B\in \mathrm{R}\text{, }&b=0\text{ and }C=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$2AC+BC=4b^{2}$$

Subtract 2AC from both sides.

$$BC=4b^{2}-2AC$$

The equation is in standard form.

$$CB=4b^{2}-2AC$$

Divide both sides by C.

$$\frac{CB}{C}=\frac{4b^{2}-2AC}{C}$$

Dividing by C undoes the multiplication by C.

$$B=\frac{4b^{2}-2AC}{C}$$

Divide -2AC+4b^{2} by C.

$$B=\frac{4b^{2}}{C}-2A$$