Solve I=2sqrt{2-3x}

$$I = 2 \sqrt { 2 - 3 x }$$

Solve for x

$$x=-\frac{I^{2}}{12}+\frac{2}{3}$$

$$I\geq 0$$

Steps for Solving Radical Equations

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

$$2\sqrt{2-3x}=I$$

Divide both sides by 2.

$$\frac{2\sqrt{-3x+2}}{2}=\frac{I}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$\sqrt{-3x+2}=\frac{I}{2}$$

Square both sides of the equation.

$$-3x+2=\frac{I^{2}}{4}$$

Subtract 2 from both sides of the equation.

$$-3x+2-2=\frac{I^{2}}{4}-2$$

Subtracting 2 from itself leaves 0.

$$-3x=\frac{I^{2}}{4}-2$$

Divide both sides by -3.

$$\frac{-3x}{-3}=\frac{\frac{I^{2}}{4}-2}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x=\frac{\frac{I^{2}}{4}-2}{-3}$$

Divide \frac{I^{2}}{4}-2 by -3.

$$x=-\frac{I^{2}}{12}+\frac{2}{3}$$

Solve for x (complex solution)

$$x=-\frac{I^{2}}{12}+\frac{2}{3}$$

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

Solve for I

$$I=2\sqrt{2-3x}$$

$$x\leq \frac{2}{3}$$