Solve 1/2(3x+2)^2-4=28

$$\frac { 1 } { 2 } ( 3 x + 2 ) ^ { 2 } - 4 = 28$$

Solve for x

$$x=2$$

$$x = -\frac{10}{3} = -3\frac{1}{3} \approx -3.333333333$$

Steps by Finding Square Root

Add 4 to both sides of the equation.

$$\frac{1}{2}\left(3x+2\right)^{2}-4+4=28+4$$

Subtracting 4 from itself leaves 0.

$$\frac{1}{2}\left(3x+2\right)^{2}=28+4$$

Add 28 to 4.

$$\frac{1}{2}\left(3x+2\right)^{2}=32$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}\left(3x+2\right)^{2}}{\frac{1}{2}}=\frac{32}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$\left(3x+2\right)^{2}=\frac{32}{\frac{1}{2}}$$

Divide 32 by \frac{1}{2} by multiplying 32 by the reciprocal of \frac{1}{2}.

$$\left(3x+2\right)^{2}=64$$

Take the square root of both sides of the equation.

$$3x+2=8 3x+2=-8$$

Subtract 2 from both sides of the equation.

$$3x+2-2=8-2 3x+2-2=-8-2$$

Subtracting 2 from itself leaves 0.

$$3x=8-2 3x=-8-2$$

Subtract 2 from 8.

$$3x=6$$

Subtract 2 from -8.

$$3x=-10$$

Divide both sides by 3.

$$\frac{3x}{3}=\frac{6}{3} \frac{3x}{3}=-\frac{10}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x=\frac{6}{3} x=-\frac{10}{3}$$

Divide 6 by 3.

$$x=2$$

Divide -10 by 3.

$$x=-\frac{10}{3}$$

The equation is now solved.

$$x=2 x=-\frac{10}{3}$$