Solve sqrt{2x-3}=y

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

Solve for x

$$x=\frac{y^{2}+3}{2}$$

$$y\geq 0$$

Steps for Solving Radical Equations

Square both sides of the equation.

$$2x-3=y^{2}$$

Add 3 to both sides of the equation.

$$2x-3-\left(-3\right)=y^{2}-\left(-3\right)$$

Subtracting -3 from itself leaves 0.

$$2x=y^{2}-\left(-3\right)$$

Subtract -3 from y^{2}.

$$2x=y^{2}+3$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{y^{2}+3}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{y^{2}+3}{2}$$

Solve for x (complex solution)

$$x=\frac{y^{2}+3}{2}$$

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

Solve for y (complex solution)

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

Solve for y

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

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