Solve 0=left(3x-2right)^2-9

$$0= { \left(3x-2 \right) }^{ 2 } -9$$

Solve for x

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

$$x=-\frac{1}{3}\approx -0.333333333$$

Steps Using Factoring By Grouping

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

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

Subtract 9 from 4 to get -5.

$$0=9x^{2}-12x-5$$

Swap sides so that all variable terms are on the left hand side.

$$9x^{2}-12x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-12 ab=9\left(-5\right)=-45$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -45.

$$1,-45 3,-15 5,-9$$

Calculate the sum for each pair.

$$1-45=-44 3-15=-12 5-9=-4$$

The solution is the pair that gives sum -12.

$$a=-15 b=3$$

Rewrite 9x^{2}-12x-5 as \left(9x^{2}-15x\right)+\left(3x-5\right).

$$\left(9x^{2}-15x\right)+\left(3x-5\right)$$

Factor out 3x in 9x^{2}-15x.

$$3x\left(3x-5\right)+3x-5$$

Factor out common term 3x-5 by using distributive property.

$$\left(3x-5\right)\left(3x+1\right)$$

To find equation solutions, solve 3x-5=0 and 3x+1=0.

$$x=\frac{5}{3} x=-\frac{1}{3}$$

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}.

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

Subtract 9 from 4 to get -5.

$$0=9x^{2}-12x-5$$

Swap sides so that all variable terms are on the left hand side.

$$9x^{2}-12x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -12 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 9\left(-5\right)}}{2\times 9}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 9\left(-5\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-12\right)±\sqrt{144-36\left(-5\right)}}{2\times 9}$$

Multiply -36 times -5.

$$x=\frac{-\left(-12\right)±\sqrt{144+180}}{2\times 9}$$

Add 144 to 180.

$$x=\frac{-\left(-12\right)±\sqrt{324}}{2\times 9}$$

Take the square root of 324.

$$x=\frac{-\left(-12\right)±18}{2\times 9}$$

The opposite of -12 is 12.

$$x=\frac{12±18}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{12±18}{18}$$

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

$$x=\frac{30}{18}$$

Reduce the fraction \frac{30}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{5}{3}$$

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

$$x=-\frac{6}{18}$$

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

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

The equation is now solved.

$$x=\frac{5}{3} x=-\frac{1}{3}$$

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}.

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

Subtract 9 from 4 to get -5.

$$0=9x^{2}-12x-5$$

Swap sides so that all variable terms are on the left hand side.

$$9x^{2}-12x-5=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$9x^{2}-12x=5$$

Divide both sides by 9.

$$\frac{9x^{2}-12x}{9}=\frac{5}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{12}{9}\right)x=\frac{5}{9}$$

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

$$x^{2}-\frac{4}{3}x=\frac{5}{9}$$

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.

$$x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=\frac{5}{9}+\left(-\frac{2}{3}\right)^{2}$$

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

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{5+4}{9}$$

Add \frac{5}{9} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=1$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

$$x=\frac{5}{3} x=-\frac{1}{3}$$