Solve (2x+1)^2=1

$$( 2 x + 1 ) ^ { 2 } = 1$$

Solve for x

$$x=0$$

$$x=-1$$

Steps by Finding Square Root

Take the square root of both sides of the equation.

$$2x+1=1 2x+1=-1$$

Subtract 1 from both sides of the equation.

$$2x+1-1=1-1 2x+1-1=-1-1$$

Subtracting 1 from itself leaves 0.

$$2x=1-1 2x=-1-1$$

Subtract 1 from 1.

$$2x=0$$

Subtract 1 from -1.

$$2x=-2$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{0}{2} \frac{2x}{2}=-\frac{2}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{0}{2} x=-\frac{2}{2}$$

Divide 0 by 2.

$$x=0$$

Divide -2 by 2.

$$x=-1$$

The equation is now solved.

$$x=0 x=-1$$