Solve y=sqrt{4x-3}-1

$$y = \sqrt { 4 x - 3 } - 1$$

Solve for x

$$x=\frac{\left(y+1\right)^{2}+3}{4}$$

$$y+1\geq 0$$

Steps for Solving Radical Equations

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

$$\sqrt{4x-3}-1=y$$

Add 1 to both sides.

$$\sqrt{4x-3}=y+1$$

Square both sides of the equation.

$$4x-3=\left(y+1\right)^{2}$$

Add 3 to both sides of the equation.

$$4x-3-\left(-3\right)=\left(y+1\right)^{2}-\left(-3\right)$$

Subtracting -3 from itself leaves 0.

$$4x=\left(y+1\right)^{2}-\left(-3\right)$$

Subtract -3 from \left(y+1\right)^{2}.

$$4x=\left(y+1\right)^{2}+3$$

Divide both sides by 4.

$$\frac{4x}{4}=\frac{\left(y+1\right)^{2}+3}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x=\frac{\left(y+1\right)^{2}+3}{4}$$

Solve for x (complex solution)

$$x=\frac{\left(y+1\right)^{2}+3}{4}$$

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

Solve for y

$$y=\sqrt{4x-3}-1$$

$$x\geq \frac{3}{4}$$