Solve (3)(3x-2)^2+(2-3x)=0

$$( 3 ) ( 3 x - 2 ) ^ { 2 } + ( 2 - 3 x ) = 0$$

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

$$x=\frac{7}{9}\approx 0.777777778$$

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}.

$$3\left(9x^{2}-12x+4\right)+2-3x=0$$

Use the distributive property to multiply 3 by 9x^{2}-12x+4.

$$27x^{2}-36x+12+2-3x=0$$

Add 12 and 2 to get 14.

$$27x^{2}-36x+14-3x=0$$

Combine -36x and -3x to get -39x.

$$27x^{2}-39x+14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 27x^{2}+ax+bx+14. To find a and b, set up a system to be solved.

$$a+b=-39 ab=27\times 14=378$$

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 378.

$$-1,-378 -2,-189 -3,-126 -6,-63 -7,-54 -9,-42 -14,-27 -18,-21$$

Calculate the sum for each pair.

$$-1-378=-379 -2-189=-191 -3-126=-129 -6-63=-69 -7-54=-61 -9-42=-51 -14-27=-41 -18-21=-39$$

The solution is the pair that gives sum -39.

$$a=-21 b=-18$$

Rewrite 27x^{2}-39x+14 as \left(27x^{2}-21x\right)+\left(-18x+14\right).

$$\left(27x^{2}-21x\right)+\left(-18x+14\right)$$

Factor out 3x in the first and -2 in the second group.

$$3x\left(9x-7\right)-2\left(9x-7\right)$$

Factor out common term 9x-7 by using distributive property.

$$\left(9x-7\right)\left(3x-2\right)$$

To find equation solutions, solve 9x-7=0 and 3x-2=0.

$$x=\frac{7}{9} x=\frac{2}{3}$$

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}.

$$3\left(9x^{2}-12x+4\right)+2-3x=0$$

Use the distributive property to multiply 3 by 9x^{2}-12x+4.

$$27x^{2}-36x+12+2-3x=0$$

Add 12 and 2 to get 14.

$$27x^{2}-36x+14-3x=0$$

Combine -36x and -3x to get -39x.

$$27x^{2}-39x+14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 27 for a, -39 for b, and 14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-39\right)±\sqrt{\left(-39\right)^{2}-4\times 27\times 14}}{2\times 27}$$

Square -39.

$$x=\frac{-\left(-39\right)±\sqrt{1521-4\times 27\times 14}}{2\times 27}$$

Multiply -4 times 27.

$$x=\frac{-\left(-39\right)±\sqrt{1521-108\times 14}}{2\times 27}$$

Multiply -108 times 14.

$$x=\frac{-\left(-39\right)±\sqrt{1521-1512}}{2\times 27}$$

Add 1521 to -1512.

$$x=\frac{-\left(-39\right)±\sqrt{9}}{2\times 27}$$

Take the square root of 9.

$$x=\frac{-\left(-39\right)±3}{2\times 27}$$

The opposite of -39 is 39.

$$x=\frac{39±3}{2\times 27}$$

Multiply 2 times 27.

$$x=\frac{39±3}{54}$$

Now solve the equation x=\frac{39±3}{54} when ± is plus. Add 39 to 3.

$$x=\frac{42}{54}$$

Reduce the fraction \frac{42}{54} to lowest terms by extracting and canceling out 6.

$$x=\frac{7}{9}$$

Now solve the equation x=\frac{39±3}{54} when ± is minus. Subtract 3 from 39.

$$x=\frac{36}{54}$$

Reduce the fraction \frac{36}{54} to lowest terms by extracting and canceling out 18.

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

The equation is now solved.

$$x=\frac{7}{9} x=\frac{2}{3}$$

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}.

$$3\left(9x^{2}-12x+4\right)+2-3x=0$$

Use the distributive property to multiply 3 by 9x^{2}-12x+4.

$$27x^{2}-36x+12+2-3x=0$$

Add 12 and 2 to get 14.

$$27x^{2}-36x+14-3x=0$$

Combine -36x and -3x to get -39x.

$$27x^{2}-39x+14=0$$

Subtract 14 from both sides. Anything subtracted from zero gives its negation.

$$27x^{2}-39x=-14$$

Divide both sides by 27.

$$\frac{27x^{2}-39x}{27}=-\frac{14}{27}$$

Dividing by 27 undoes the multiplication by 27.

$$x^{2}+\left(-\frac{39}{27}\right)x=-\frac{14}{27}$$

Reduce the fraction \frac{-39}{27} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{13}{9}x=-\frac{14}{27}$$

Divide -\frac{13}{9}, the coefficient of the x term, by 2 to get -\frac{13}{18}. Then add the square of -\frac{13}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{9}x+\left(-\frac{13}{18}\right)^{2}=-\frac{14}{27}+\left(-\frac{13}{18}\right)^{2}$$

Square -\frac{13}{18} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{9}x+\frac{169}{324}=-\frac{14}{27}+\frac{169}{324}$$

Add -\frac{14}{27} to \frac{169}{324} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{13}{9}x+\frac{169}{324}=\frac{1}{324}$$

Factor x^{2}-\frac{13}{9}x+\frac{169}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{13}{18}\right)^{2}=\frac{1}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{18}\right)^{2}}=\sqrt{\frac{1}{324}}$$

Simplify.

$$x-\frac{13}{18}=\frac{1}{18} x-\frac{13}{18}=-\frac{1}{18}$$

Add \frac{13}{18} to both sides of the equation.

$$x=\frac{7}{9} x=\frac{2}{3}$$