Solve (4n+1)^2=12

$$( 4 n + 1 ) ^ { 2 } = 12$$

Solve for n

$$n=\frac{\sqrt{3}}{2}-\frac{1}{4}\approx 0.616025404$$

$$n=-\frac{\sqrt{3}}{2}-\frac{1}{4}\approx -1.116025404$$

Steps by Finding Square Root

Take the square root of both sides of the equation.

$$4n+1=2\sqrt{3} 4n+1=-2\sqrt{3}$$

Subtract 1 from both sides of the equation.

$$4n+1-1=2\sqrt{3}-1 4n+1-1=-2\sqrt{3}-1$$

Subtracting 1 from itself leaves 0.

$$4n=2\sqrt{3}-1 4n=-2\sqrt{3}-1$$

Subtract 1 from 2\sqrt{3}.

$$4n=2\sqrt{3}-1$$

Subtract 1 from -2\sqrt{3}.

$$4n=-2\sqrt{3}-1$$

Divide both sides by 4.

$$\frac{4n}{4}=\frac{2\sqrt{3}-1}{4} \frac{4n}{4}=\frac{-2\sqrt{3}-1}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$n=\frac{2\sqrt{3}-1}{4} n=\frac{-2\sqrt{3}-1}{4}$$

Divide 2\sqrt{3}-1 by 4.

$$n=\frac{\sqrt{3}}{2}-\frac{1}{4}$$

Divide -2\sqrt{3}-1 by 4.

$$n=-\frac{\sqrt{3}}{2}-\frac{1}{4}$$

The equation is now solved.

$$n=\frac{\sqrt{3}}{2}-\frac{1}{4} n=-\frac{\sqrt{3}}{2}-\frac{1}{4}$$