Solve K=3sqrt{2r+st}

$$K = 3 \sqrt { 2 r + s t }$$

Solve for r

$$r=\frac{K^{2}-9st}{18}$$

$$K\geq 0$$

Steps for Solving Radical Equations

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

$$3\sqrt{2r+st}=K$$

Divide both sides by 3.

$$\frac{3\sqrt{2r+st}}{3}=\frac{K}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$\sqrt{2r+st}=\frac{K}{3}$$

Square both sides of the equation.

$$2r+st=\frac{K^{2}}{9}$$

Subtract st from both sides of the equation.

$$2r+st-st=\frac{K^{2}}{9}-st$$

Subtracting st from itself leaves 0.

$$2r=\frac{K^{2}}{9}-st$$

Subtract st from \frac{K^{2}}{9}.

$$2r=-st+\frac{K^{2}}{9}$$

Divide both sides by 2.

$$\frac{2r}{2}=\frac{-st+\frac{K^{2}}{9}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$r=\frac{-st+\frac{K^{2}}{9}}{2}$$

Divide \frac{K^{2}}{9}-st by 2.

$$r=-\frac{st}{2}+\frac{K^{2}}{18}$$

Solve for r (complex solution)

$$r=\frac{K^{2}-9st}{18}$$

$$arg(K)<\pi \text{ or }K=0$$

Solve for K

$$K=3\sqrt{2r+st}$$

$$r\geq -\frac{st}{2}$$