Solve (x+B/2A)^2=(frac{sqrt{Delta}}{2A})^2

$$( x + \frac { B } { 2 A } ) ^ { 2 } = ( \frac { \sqrt { \Delta } } { 2 A } ) ^ { 2 }$$

Solve for A

$$\left\{\begin{matrix}A=-\frac{B-\sqrt{\Delta }}{2x}\text{, }&B\neq \sqrt{\Delta }\text{ and }x\neq 0\text{ and }\Delta \geq 0\text{ and }\Delta \neq B^{2}\\A=-\frac{B+\sqrt{\Delta }}{2x}\text{, }&B\neq -\sqrt{\Delta }\text{ and }x\neq 0\text{ and }\Delta \geq 0\text{ and }\Delta \neq B^{2}\\A=-\frac{B}{x}\text{, }&B\neq 0\text{ and }x\neq 0\text{ and }\Delta =B^{2}\\A\neq 0\text{, }&\left(\Delta =0\text{ and }B=0\text{ and }x=0\right)\text{ or }\left(\Delta =B^{2}\text{ and }x=0\right)\end{matrix}\right.$$

Solve for B

$$B=-\left(2Ax+\sqrt{\Delta }\right)$$

$$B=-2Ax+\sqrt{\Delta }\text{, }A\neq 0\text{ and }\Delta \geq 0$$