Solve (3)(2x+1)^2-5=0

$$( 3 ) ( 2 x + 1 ) ^ { 2 } - 5 = 0$$

Solve for x

$$x=\frac{\sqrt{15}}{6}-\frac{1}{2}\approx 0.145497224$$

$$x=-\frac{\sqrt{15}}{6}-\frac{1}{2}\approx -1.145497224$$

Steps by Finding Square Root

Add 5 to both sides of the equation.

$$3\left(2x+1\right)^{2}-5+5=5$$

Subtracting 5 from itself leaves 0.

$$3\left(2x+1\right)^{2}=5$$

Divide both sides by 3.

$$\frac{3\left(2x+1\right)^{2}}{3}=\frac{5}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$\left(2x+1\right)^{2}=\frac{5}{3}$$

Take the square root of both sides of the equation.

$$2x+1=\frac{\sqrt{15}}{3} 2x+1=-\frac{\sqrt{15}}{3}$$

Subtract 1 from both sides of the equation.

$$2x+1-1=\frac{\sqrt{15}}{3}-1 2x+1-1=-\frac{\sqrt{15}}{3}-1$$

Subtracting 1 from itself leaves 0.

$$2x=\frac{\sqrt{15}}{3}-1 2x=-\frac{\sqrt{15}}{3}-1$$

Subtract 1 from \frac{\sqrt{15}}{3}.

$$2x=\frac{\sqrt{15}}{3}-1$$

Subtract 1 from -\frac{\sqrt{15}}{3}.

$$2x=-\frac{\sqrt{15}}{3}-1$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{\frac{\sqrt{15}}{3}-1}{2} \frac{2x}{2}=\frac{-\frac{\sqrt{15}}{3}-1}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{\frac{\sqrt{15}}{3}-1}{2} x=\frac{-\frac{\sqrt{15}}{3}-1}{2}$$

Divide \frac{\sqrt{15}}{3}-1 by 2.

$$x=\frac{\sqrt{15}}{6}-\frac{1}{2}$$

Divide -\frac{\sqrt{15}}{3}-1 by 2.

$$x=-\frac{\sqrt{15}}{6}-\frac{1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{15}}{6}-\frac{1}{2} x=-\frac{\sqrt{15}}{6}-\frac{1}{2}$$