Solve v=sqrt{b^2+2gs}

$$v = \sqrt { b ^ { 2 } + 2 g s }$$

Solve for g

$$\left\{\begin{matrix}g=-\frac{b^{2}-v^{2}}{2s}\text{, }&\left(v\geq 0\text{ and }s\neq 0\text{ and }v>-b\text{ and }v\leq b\right)\text{ or }\left(s\neq 0\text{ and }v=-b\text{ and }b\leq 0\right)\text{ or }\left(v\geq 0\text{ and }s\neq 0\text{ and }b\leq -\sqrt{\left(b-v\right)\left(v+b\right)}\text{ and }v\leq -b\text{ and }v\geq b\right)\text{ or }\left(v\geq 0\text{ and }s\neq 0\text{ and }b\leq -\sqrt{\left(b-v\right)\left(v+b\right)}\text{ and }v\leq b\text{ and }v\geq -b\right)\text{ or }\left(s\neq 0\text{ and }v>|b|\right)\\g\in \mathrm{R}\text{, }&v=|b|\text{ and }s=0\end{matrix}\right.$$

Steps for Solving Radical Equations

Swap sides so that all variable terms are on the left hand side.

$$\sqrt{b^{2}+2gs}=v$$

Square both sides of the equation.

$$2sg+b^{2}=v^{2}$$

Subtract b^{2} from both sides of the equation.

$$2sg+b^{2}-b^{2}=v^{2}-b^{2}$$

Subtracting b^{2} from itself leaves 0.

$$2sg=v^{2}-b^{2}$$

Subtract b^{2} from v^{2}.

$$2sg=\left(v-b\right)\left(v+b\right)$$

Divide both sides by 2s.

$$\frac{2sg}{2s}=\frac{\left(v-b\right)\left(v+b\right)}{2s}$$

Dividing by 2s undoes the multiplication by 2s.

$$g=\frac{\left(v-b\right)\left(v+b\right)}{2s}$$

Solve for b (complex solution)

$$b=-\sqrt{v^{2}-2gs}$$

$$b=\sqrt{v^{2}-2gs}\text{, }arg(v)<\pi \text{ or }v=0$$

Solve for g (complex solution)

$$\left\{\begin{matrix}g=-\frac{b^{2}-v^{2}}{2s}\text{, }&s\neq 0\text{ and }\left(arg(v)<\pi \text{ or }v=0\right)\\g\in \mathrm{C}\text{, }&\left(v=0\text{ and }s=0\text{ and }b=0\right)\text{ or }\left(v=-b\text{ and }s=0\text{ and }arg(b)\geq \pi \text{ and }b\neq 0\right)\text{ or }\left(v=b\text{ and }s=0\text{ and }arg(b)<\pi \text{ and }b\neq 0\right)\end{matrix}\right.$$