Solve (2x+1)^2-5=0

$$( 2 x + 1 ) ^ { 2 } - 5 = 0$$

Solve for x

$$x=\frac{\sqrt{5}-1}{2}\approx 0.618033989$$

$$x=\frac{-\sqrt{5}-1}{2}\approx -1.618033989$$

Steps by Finding Square Root

Add 5 to both sides of the equation.

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

Subtracting 5 from itself leaves 0.

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

Take the square root of both sides of the equation.

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

Subtract 1 from both sides of the equation.

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

Subtracting 1 from itself leaves 0.

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

Subtract 1 from \sqrt{5}.

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

Subtract 1 from -\sqrt{5}.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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