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

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

Solve for x

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

$$y-1\geq 0$$

Steps for Solving Radical Equations

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

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

Subtract 1 from both sides.

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

Square both sides of the equation.

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

Subtract 2 from both sides of the equation.

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

Subtracting 2 from itself leaves 0.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide \left(y-1\right)^{2}-2 by 2.

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

Solve for x (complex solution)

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

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

Solve for y (complex solution)

$$y=\sqrt{2\left(x+1\right)}+1$$

Solve for y

$$y=\sqrt{2\left(x+1\right)}+1$$

$$x\geq -1$$