Solve AC^2=2AB^2+BC^2

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

Solve for A

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

Steps for Solving Linear Equation

Subtract 2AB^{2} from both sides.

$$AC^{2}-2AB^{2}=BC^{2}$$

Combine all terms containing A.

$$\left(C^{2}-2B^{2}\right)A=BC^{2}$$

Divide both sides by -2B^{2}+C^{2}.

$$\frac{\left(C^{2}-2B^{2}\right)A}{C^{2}-2B^{2}}=\frac{BC^{2}}{C^{2}-2B^{2}}$$

Dividing by -2B^{2}+C^{2} undoes the multiplication by -2B^{2}+C^{2}.

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

Solve for B

$$\left\{\begin{matrix}B=-\frac{C\left(\sqrt{8A^{2}+C^{2}}+C\right)}{4A}\text{; }B=-\frac{C\left(-\sqrt{8A^{2}+C^{2}}+C\right)}{4A}\text{, }&A\neq 0\\B=0\text{, }&A=0\text{ and }C\neq 0\\B\in \mathrm{R}\text{, }&A=0\text{ and }C=0\end{matrix}\right.$$