Solve sqrt{a}=x-b^2/2c

$$\sqrt { a } = \frac { x - b ^ { 2 } } { 2 c }$$

Solve for a

$$a=\frac{\left(\frac{x-b^{2}}{c}\right)^{2}}{4}$$

$$\left(c>0\text{ or }x\leq b^{2}\right)\text{ and }\left(c<0\text{ or }x\geq b^{2}\right)\text{ and }c\neq 0$$

Steps for Solving Radical Equations

Square both sides of the equation.

$$a=\frac{\left(x-b^{2}\right)^{2}}{4c^{2}}$$

Solve for a (complex solution)

$$a=\frac{\left(\frac{x-b^{2}}{c}\right)^{2}}{4}$$

$$\left(x=b^{2}\text{ or }arg(\frac{x-b^{2}}{c})<\pi \right)\text{ and }c\neq 0$$

Solve for b (complex solution)

$$b=-\sqrt{-2\sqrt{a}c+x}$$

$$b=\sqrt{-2\sqrt{a}c+x}\text{, }c\neq 0$$

Solve for b

$$b=\sqrt{-2\sqrt{a}c+x}$$

$$b=-\sqrt{-2\sqrt{a}c+x}\text{, }\left(x>0\text{ and }a\geq 0\text{ and }c<0\right)\text{ or }\left(x>0\text{ and }a\leq \frac{x^{2}}{4c^{2}}\text{ and }a\geq 0\text{ and }c\neq 0\right)\text{ or }\left(x\geq 0\text{ and }c<0\text{ and }a\geq \frac{x^{2}}{4c^{2}}\right)\text{ or }\left(a>0\text{ and }c<0\text{ and }a\geq \frac{x^{2}}{4c^{2}}\right)\text{ or }\left(x\geq 0\text{ and }c\neq 0\text{ and }a=0\right)$$