Solve (3x-2)^2-4=9x-6

$$( 3 x - 2 ) ^ { 2 } - 4 = 9 x - 6$$

Solve for x

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

$$x=2$$

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}.

$$9x^{2}-12x+4-4=9x-6$$

Subtract 4 from 4 to get 0.

$$9x^{2}-12x=9x-6$$

Subtract 9x from both sides.

$$9x^{2}-12x-9x=-6$$

Combine -12x and -9x to get -21x.

$$9x^{2}-21x=-6$$

Add 6 to both sides.

$$9x^{2}-21x+6=0$$

Divide both sides by 3.

$$3x^{2}-7x+2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=-7 ab=3\times 2=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -7.

$$a=-6 b=-1$$

Rewrite 3x^{2}-7x+2 as \left(3x^{2}-6x\right)+\left(-x+2\right).

$$\left(3x^{2}-6x\right)+\left(-x+2\right)$$

Factor out 3x in the first and -1 in the second group.

$$3x\left(x-2\right)-\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(3x-1\right)$$

To find equation solutions, solve x-2=0 and 3x-1=0.

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

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}.

$$9x^{2}-12x+4-4=9x-6$$

Subtract 4 from 4 to get 0.

$$9x^{2}-12x=9x-6$$

Subtract 9x from both sides.

$$9x^{2}-12x-9x=-6$$

Combine -12x and -9x to get -21x.

$$9x^{2}-21x=-6$$

Add 6 to both sides.

$$9x^{2}-21x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -21 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 9\times 6}}{2\times 9}$$

Square -21.

$$x=\frac{-\left(-21\right)±\sqrt{441-4\times 9\times 6}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-21\right)±\sqrt{441-36\times 6}}{2\times 9}$$

Multiply -36 times 6.

$$x=\frac{-\left(-21\right)±\sqrt{441-216}}{2\times 9}$$

Add 441 to -216.

$$x=\frac{-\left(-21\right)±\sqrt{225}}{2\times 9}$$

Take the square root of 225.

$$x=\frac{-\left(-21\right)±15}{2\times 9}$$

The opposite of -21 is 21.

$$x=\frac{21±15}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{21±15}{18}$$

Now solve the equation x=\frac{21±15}{18} when ± is plus. Add 21 to 15.

$$x=\frac{36}{18}$$

Divide 36 by 18.

$$x=2$$

Now solve the equation x=\frac{21±15}{18} when ± is minus. Subtract 15 from 21.

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

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

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

The equation is now solved.

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

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}.

$$9x^{2}-12x+4-4=9x-6$$

Subtract 4 from 4 to get 0.

$$9x^{2}-12x=9x-6$$

Subtract 9x from both sides.

$$9x^{2}-12x-9x=-6$$

Combine -12x and -9x to get -21x.

$$9x^{2}-21x=-6$$

Divide both sides by 9.

$$\frac{9x^{2}-21x}{9}=-\frac{6}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{21}{9}\right)x=-\frac{6}{9}$$

Reduce the fraction \frac{-21}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{7}{3}x=-\frac{6}{9}$$

Reduce the fraction \frac{-6}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{7}{3}x=-\frac{2}{3}$$

Divide -\frac{7}{3}, the coefficient of the x term, by 2 to get -\frac{7}{6}. Then add the square of -\frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{3}x+\left(-\frac{7}{6}\right)^{2}=-\frac{2}{3}+\left(-\frac{7}{6}\right)^{2}$$

Square -\frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=-\frac{2}{3}+\frac{49}{36}$$

Add -\frac{2}{3} to \frac{49}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=\frac{25}{36}$$

Factor x^{2}-\frac{7}{3}x+\frac{49}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{6}\right)^{2}=\frac{25}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Simplify.

$$x-\frac{7}{6}=\frac{5}{6} x-\frac{7}{6}=-\frac{5}{6}$$

Add \frac{7}{6} to both sides of the equation.

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