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

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

Solve for x

$$x=\frac{\left(y+1\right)^{2}+32}{16}$$

$$\frac{y+1}{4}\geq 0$$

Steps for Solving Radical Equations

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

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

Add 1 to both sides.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Square both sides of the equation.

$$x-2=\frac{\left(y+1\right)^{2}}{16}$$

Add 2 to both sides of the equation.

$$x-2-\left(-2\right)=\frac{\left(y+1\right)^{2}}{16}-\left(-2\right)$$

Subtracting -2 from itself leaves 0.

$$x=\frac{\left(y+1\right)^{2}}{16}-\left(-2\right)$$

Subtract -2 from \frac{\left(y+1\right)^{2}}{16}.

$$x=\frac{\left(y+1\right)^{2}}{16}+2$$

Solve for x (complex solution)

$$x=\frac{\left(y+1\right)^{2}+32}{16}$$

$$y=-1\text{ or }arg(\frac{y+1}{4})<\pi$$

Solve for y

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

$$x\geq 2$$