Solve (3x-2)^2=4

$$( 3 x - 2 ) ^ { 2 } = 4$$

Solve for x

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

$$x=0$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-2\right)^{2}.

$$9x^{2}-12x+4=4$$

Subtract 4 from both sides.

$$9x^{2}-12x+4-4=0$$

Subtract 4 from 4 to get 0.

$$9x^{2}-12x=0$$

Factor out x.

$$x\left(9x-12\right)=0$$

To find equation solutions, solve x=0 and 9x-12=0.

$$x=0 x=\frac{4}{3}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-2\right)^{2}.

$$9x^{2}-12x+4=4$$

Subtract 4 from both sides.

$$9x^{2}-12x+4-4=0$$

Subtract 4 from 4 to get 0.

$$9x^{2}-12x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -12 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}}}{2\times 9}$$

Take the square root of \left(-12\right)^{2}.

$$x=\frac{-\left(-12\right)±12}{2\times 9}$$

The opposite of -12 is 12.

$$x=\frac{12±12}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{12±12}{18}$$

Now solve the equation x=\frac{12±12}{18} when ± is plus. Add 12 to 12.

$$x=\frac{24}{18}$$

Reduce the fraction \frac{24}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{12±12}{18} when ± is minus. Subtract 12 from 12.

$$x=\frac{0}{18}$$

Divide 0 by 18.

$$x=0$$

The equation is now solved.

$$x=\frac{4}{3} x=0$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-2\right)^{2}.

$$9x^{2}-12x+4=4$$

Subtract 4 from both sides.

$$9x^{2}-12x=4-4$$

Subtract 4 from 4 to get 0.

$$9x^{2}-12x=0$$

Divide both sides by 9.

$$\frac{9x^{2}-12x}{9}=\frac{0}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{12}{9}\right)x=\frac{0}{9}$$

Reduce the fraction \frac{-12}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{4}{3}x=\frac{0}{9}$$

Divide 0 by 9.

$$x^{2}-\frac{4}{3}x=0$$

Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{4}{9}$$

Factor x^{2}-\frac{4}{3}x+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

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

Add \frac{2}{3} to both sides of the equation.

$$x=\frac{4}{3} x=0$$