Solve c=sqrt{4h(2R-h)}

$$c = \sqrt { 4 h ( 2 R - h ) }$$

Solve for R

$$\left\{\begin{matrix}R=\frac{c^{2}}{8h}+\frac{h}{2}\text{, }&c\geq 0\text{ and }h\neq 0\\R\in \mathrm{R}\text{, }&h=0\text{ and }c=0\end{matrix}\right.$$

Steps for Solving Radical Equations

Use the distributive property to multiply 4h by 2R-h.

$$c=\sqrt{8hR-4h^{2}}$$

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

$$\sqrt{8hR-4h^{2}}=c$$

Square both sides of the equation.

$$8hR-4h^{2}=c^{2}$$

Subtract -4h^{2} from both sides of the equation.

$$8hR-4h^{2}-\left(-4h^{2}\right)=c^{2}-\left(-4h^{2}\right)$$

Subtracting -4h^{2} from itself leaves 0.

$$8hR=c^{2}-\left(-4h^{2}\right)$$

Subtract -4h^{2} from c^{2}.

$$8hR=c^{2}+4h^{2}$$

Divide both sides by 8h.

$$\frac{8hR}{8h}=\frac{c^{2}+4h^{2}}{8h}$$

Dividing by 8h undoes the multiplication by 8h.

$$R=\frac{c^{2}+4h^{2}}{8h}$$

Divide c^{2}+4h^{2} by 8h.

$$R=\frac{c^{2}}{8h}+\frac{h}{2}$$

Solve for R (complex solution)

$$\left\{\begin{matrix}R=\frac{c^{2}}{8h}+\frac{h}{2}\text{, }&h\neq 0\text{ and }\left(c=0\text{ or }arg(c)<\pi \right)\\R\in \mathrm{C}\text{, }&c=0\text{ and }h=0\end{matrix}\right.$$

Solve for c (complex solution)

$$c=2\sqrt{h\left(2R-h\right)}$$

Solve for c

$$c=2\sqrt{h\left(2R-h\right)}$$

$$\left(h\leq 0\text{ or }h\leq 2R\right)\text{ and }\left(h\geq 0\text{ or }h\geq 2R\right)$$