Solve left(3x+1right)^2=3

$${ \left(3x+1 \right) }^{ 2 } =3$$

Solve for x

$$x=\frac{\sqrt{3}-1}{3}\approx 0.244016936$$

$$x=\frac{-\sqrt{3}-1}{3}\approx -0.910683603$$

Steps by Finding Square Root

Take the square root of both sides of the equation.

$$3x+1=\sqrt{3} 3x+1=-\sqrt{3}$$

Subtract 1 from both sides of the equation.

$$3x+1-1=\sqrt{3}-1 3x+1-1=-\sqrt{3}-1$$

Subtracting 1 from itself leaves 0.

$$3x=\sqrt{3}-1 3x=-\sqrt{3}-1$$

Subtract 1 from \sqrt{3}.

$$3x=\sqrt{3}-1$$

Subtract 1 from -\sqrt{3}.

$$3x=-\sqrt{3}-1$$

Divide both sides by 3.

$$\frac{3x}{3}=\frac{\sqrt{3}-1}{3} \frac{3x}{3}=\frac{-\sqrt{3}-1}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x=\frac{\sqrt{3}-1}{3} x=\frac{-\sqrt{3}-1}{3}$$