Solve 6-sqrt{4+3x}=2

$$6 - \sqrt { 4 + 3 x } = 2$$

Solve for x

$$x=4$$

Steps for Solving Radical Equations

Subtract 6 from both sides of the equation.

$$-\sqrt{3x+4}+6-6=2-6$$

Subtracting 6 from itself leaves 0.

$$-\sqrt{3x+4}=2-6$$

Subtract 6 from 2.

$$-\sqrt{3x+4}=-4$$

Divide both sides by -1.

$$\frac{-\sqrt{3x+4}}{-1}=-\frac{4}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$\sqrt{3x+4}=-\frac{4}{-1}$$

Divide -4 by -1.

$$\sqrt{3x+4}=4$$

Square both sides of the equation.

$$3x+4=16$$

Subtract 4 from both sides of the equation.

$$3x+4-4=16-4$$

Subtracting 4 from itself leaves 0.

$$3x=16-4$$

Subtract 4 from 16.

$$3x=12$$

Divide both sides by 3.

$$\frac{3x}{3}=\frac{12}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x=\frac{12}{3}$$

Divide 12 by 3.

$$x=4$$