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

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

Solve for x

$$x=-\frac{\left(1-y\right)\left(y-3\right)}{2}$$

$$2-y\geq 0$$

Steps for Solving Radical Equations

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

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

Subtract 2 from both sides.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide y-2 by -1.

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

Square both sides of the equation.

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

Subtract 1 from both sides of the equation.

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

Subtracting 1 from itself leaves 0.

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

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

$$2x=y^{2}-4y+3$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{y^{2}-4y+3}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{y^{2}-4y+3}{2}$$

Divide y^{2}-4y+3 by 2.

$$x=\frac{\left(y-3\right)\left(y-1\right)}{2}$$

Solve for x (complex solution)

$$x=-\frac{\left(1-y\right)\left(y-3\right)}{2}$$

$$y=2\text{ or }arg(2-y)<\pi$$

Solve for y (complex solution)

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

Solve for y

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

$$x\geq -\frac{1}{2}$$