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

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

Solve for A

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

Steps for Solving Linear Equation

Subtract AB^{2} from both sides.

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

Reorder the terms.

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

Combine all terms containing A.

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

The equation is in standard form.

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

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

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

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

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

Solve for B

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