Solve y=3-sqrt{x-4}

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

Solve for x

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

$$3-y\geq 0$$

Steps for Solving Radical Equations

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

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

Subtract 3 from both sides.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide y-3 by -1.

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

Square both sides of the equation.

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

Add 4 to both sides of the equation.

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

Subtracting -4 from itself leaves 0.

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

Subtract -4 from \left(-y+3\right)^{2}.

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

Solve for x (complex solution)

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

$$y=3\text{ or }arg(3-y)<\pi$$

Solve for y (complex solution)

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

Solve for y

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

$$x\geq 4$$