Solve 4AB^2=AC^2+BD^2

$$4 AB ^ { 2 } = AC ^ { 2 } + BD ^ { 2 }$$

Solve for A

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

Steps for Solving Linear Equation

Subtract AC^{2} from both sides.

$$4AB^{2}-AC^{2}=BD^{2}$$

Combine all terms containing A.

$$\left(4B^{2}-C^{2}\right)A=BD^{2}$$

Divide both sides by 4B^{2}-C^{2}.

$$\frac{\left(4B^{2}-C^{2}\right)A}{4B^{2}-C^{2}}=\frac{BD^{2}}{4B^{2}-C^{2}}$$

Dividing by 4B^{2}-C^{2} undoes the multiplication by 4B^{2}-C^{2}.

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

Solve for B

$$\left\{\begin{matrix}B=\frac{\sqrt{D^{4}+16\left(AC\right)^{2}}+D^{2}}{8A}\text{; }B=\frac{-\sqrt{D^{4}+16\left(AC\right)^{2}}+D^{2}}{8A}\text{, }&A\neq 0\\B=0\text{, }&A=0\text{ and }D\neq 0\\B\in \mathrm{R}\text{, }&A=0\text{ and }D=0\end{matrix}\right.$$