Solve Y=sqrt{x-4}+3

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

Solve for x

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

$$-\left(3-Y\right)\geq 0$$

Steps for Solving Radical Equations

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

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

Subtract 3 from both sides.

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

Square both sides of the equation.

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

Add 4 to both sides of the equation.

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

Subtracting -4 from itself leaves 0.

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

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

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

Solve for x (complex solution)

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

$$Y=3\text{ or }arg(3-Y)\geq \pi$$

Solve for Y

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

$$x\geq 4$$