Solve y=sqrt{2x-5}+1

$$y = \sqrt { 2 x - 5 } + 1$$

Solve for x

$$x=\frac{y^{2}-2y+6}{2}$$

$$y-1\geq 0$$

Steps for Solving Radical Equations

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

$$\sqrt{2x-5}+1=y$$

Subtract 1 from both sides.

$$\sqrt{2x-5}=y-1$$

Square both sides of the equation.

$$2x-5=\left(y-1\right)^{2}$$

Add 5 to both sides of the equation.

$$2x-5-\left(-5\right)=\left(y-1\right)^{2}-\left(-5\right)$$

Subtracting -5 from itself leaves 0.

$$2x=\left(y-1\right)^{2}-\left(-5\right)$$

Subtract -5 from \left(y-1\right)^{2}.

$$2x=\left(y-1\right)^{2}+5$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{\left(y-1\right)^{2}+5}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{\left(y-1\right)^{2}+5}{2}$$

Solve for x (complex solution)

$$x=\frac{y^{2}-2y+6}{2}$$

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

Solve for y

$$y=\sqrt{2x-5}+1$$

$$x\geq \frac{5}{2}$$