Solve 2w=sqrt{2u+v}(v)

$$2 w = \sqrt { 2 u + v } ( v )$$

Solve for u

$$\left\{\begin{matrix}u=-\frac{v}{2}+2\times \left(\frac{w}{v}\right)^{2}\text{, }&\left(w\geq 0\text{ and }v>0\right)\text{ or }\left(w\leq 0\text{ and }v<0\right)\\u\geq 0\text{, }&w=0\text{ and }v=0\end{matrix}\right.$$

Steps for Solving Radical Equations

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

$$\sqrt{2u+v}v=2w$$

Divide both sides by v.

$$\frac{v\sqrt{2u+v}}{v}=\frac{2w}{v}$$

Dividing by v undoes the multiplication by v.

$$\sqrt{2u+v}=\frac{2w}{v}$$

Square both sides of the equation.

$$2u+v=\frac{4w^{2}}{v^{2}}$$

Subtract v from both sides of the equation.

$$2u+v-v=\frac{4w^{2}}{v^{2}}-v$$

Subtracting v from itself leaves 0.

$$2u=\frac{4w^{2}}{v^{2}}-v$$

Subtract v from \frac{4w^{2}}{v^{2}}.

$$2u=-v+\frac{4w^{2}}{v^{2}}$$

Divide both sides by 2.

$$\frac{2u}{2}=\frac{-v+\frac{4w^{2}}{v^{2}}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$u=\frac{-v+\frac{4w^{2}}{v^{2}}}{2}$$

Divide -v+\frac{4w^{2}}{v^{2}} by 2.

$$u=-\frac{v}{2}+\frac{2w^{2}}{v^{2}}$$