Solve y=sqrt{2x-4}-1

$$y = \sqrt{ 2x-4 } -1$$

Solve for x

$$x=\frac{\left(y+1\right)^{2}+4}{2}$$

$$y+1\geq 0$$

Steps for Solving Radical Equations

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

$$\sqrt{2x-4}-1=y$$

Add 1 to both sides.

$$\sqrt{2x-4}=y+1$$

Square both sides of the equation.

$$2x-4=\left(y+1\right)^{2}$$

Add 4 to both sides of the equation.

$$2x-4-\left(-4\right)=\left(y+1\right)^{2}-\left(-4\right)$$

Subtracting -4 from itself leaves 0.

$$2x=\left(y+1\right)^{2}-\left(-4\right)$$

Subtract -4 from \left(y+1\right)^{2}.

$$2x=\left(y+1\right)^{2}+4$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{\left(y+1\right)^{2}+4}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{\left(y+1\right)^{2}+4}{2}$$

Divide \left(y+1\right)^{2}+4 by 2.

$$x=\frac{\left(y+1\right)^{2}}{2}+2$$

Solve for x (complex solution)

$$x=\frac{\left(y+1\right)^{2}+4}{2}$$

$$y=-1\text{ or }arg(y+1)<\pi$$

Solve for y (complex solution)

$$y=\sqrt{2\left(x-2\right)}-1$$

Solve for y

$$y=\sqrt{2\left(x-2\right)}-1$$

$$x\geq 2$$