Solve left(6x+2right)^2+4=28

$${ \left(6x+2 \right) }^{ 2 } +4=28$$

Solve for x

$$x=\frac{\sqrt{6}-1}{3}\approx 0.483163248$$

$$x=\frac{-\sqrt{6}-1}{3}\approx -1.149829914$$

Steps by Finding Square Root

Subtract 4 from both sides of the equation.

$$\left(6x+2\right)^{2}+4-4=28-4$$

Subtracting 4 from itself leaves 0.

$$\left(6x+2\right)^{2}=28-4$$

Subtract 4 from 28.

$$\left(6x+2\right)^{2}=24$$

Take the square root of both sides of the equation.

$$6x+2=2\sqrt{6} 6x+2=-2\sqrt{6}$$

Subtract 2 from both sides of the equation.

$$6x+2-2=2\sqrt{6}-2 6x+2-2=-2\sqrt{6}-2$$

Subtracting 2 from itself leaves 0.

$$6x=2\sqrt{6}-2 6x=-2\sqrt{6}-2$$

Subtract 2 from 2\sqrt{6}.

$$6x=2\sqrt{6}-2$$

Subtract 2 from -2\sqrt{6}.

$$6x=-2\sqrt{6}-2$$

Divide both sides by 6.

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

Dividing by 6 undoes the multiplication by 6.

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

Divide 2\sqrt{6}-2 by 6.

$$x=\frac{\sqrt{6}-1}{3}$$

Divide -2\sqrt{6}-2 by 6.

$$x=\frac{-\sqrt{6}-1}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{6}-1}{3} x=\frac{-\sqrt{6}-1}{3}$$