Solve left(3x-4right)^2=9x-12

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

Solve for x

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

$$x = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-4\right)^{2}.

$$9x^{2}-24x+16=9x-12$$

Subtract 9x from both sides.

$$9x^{2}-24x+16-9x=-12$$

Combine -24x and -9x to get -33x.

$$9x^{2}-33x+16=-12$$

Add 12 to both sides.

$$9x^{2}-33x+16+12=0$$

Add 16 and 12 to get 28.

$$9x^{2}-33x+28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx+28. To find a and b, set up a system to be solved.

$$a+b=-33 ab=9\times 28=252$$

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 252.

$$-1,-252 -2,-126 -3,-84 -4,-63 -6,-42 -7,-36 -9,-28 -12,-21 -14,-18$$

Calculate the sum for each pair.

$$-1-252=-253 -2-126=-128 -3-84=-87 -4-63=-67 -6-42=-48 -7-36=-43 -9-28=-37 -12-21=-33 -14-18=-32$$

The solution is the pair that gives sum -33.

$$a=-21 b=-12$$

Rewrite 9x^{2}-33x+28 as \left(9x^{2}-21x\right)+\left(-12x+28\right).

$$\left(9x^{2}-21x\right)+\left(-12x+28\right)$$

Factor out 3x in the first and -4 in the second group.

$$3x\left(3x-7\right)-4\left(3x-7\right)$$

Factor out common term 3x-7 by using distributive property.

$$\left(3x-7\right)\left(3x-4\right)$$

To find equation solutions, solve 3x-7=0 and 3x-4=0.

$$x=\frac{7}{3} x=\frac{4}{3}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-4\right)^{2}.

$$9x^{2}-24x+16=9x-12$$

Subtract 9x from both sides.

$$9x^{2}-24x+16-9x=-12$$

Combine -24x and -9x to get -33x.

$$9x^{2}-33x+16=-12$$

Add 12 to both sides.

$$9x^{2}-33x+16+12=0$$

Add 16 and 12 to get 28.

$$9x^{2}-33x+28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -33 for b, and 28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-33\right)±\sqrt{\left(-33\right)^{2}-4\times 9\times 28}}{2\times 9}$$

Square -33.

$$x=\frac{-\left(-33\right)±\sqrt{1089-4\times 9\times 28}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-33\right)±\sqrt{1089-36\times 28}}{2\times 9}$$

Multiply -36 times 28.

$$x=\frac{-\left(-33\right)±\sqrt{1089-1008}}{2\times 9}$$

Add 1089 to -1008.

$$x=\frac{-\left(-33\right)±\sqrt{81}}{2\times 9}$$

Take the square root of 81.

$$x=\frac{-\left(-33\right)±9}{2\times 9}$$

The opposite of -33 is 33.

$$x=\frac{33±9}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{33±9}{18}$$

Now solve the equation x=\frac{33±9}{18} when ± is plus. Add 33 to 9.

$$x=\frac{42}{18}$$

Reduce the fraction \frac{42}{18} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{33±9}{18} when ± is minus. Subtract 9 from 33.

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

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

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

The equation is now solved.

$$x=\frac{7}{3} x=\frac{4}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-4\right)^{2}.

$$9x^{2}-24x+16=9x-12$$

Subtract 9x from both sides.

$$9x^{2}-24x+16-9x=-12$$

Combine -24x and -9x to get -33x.

$$9x^{2}-33x+16=-12$$

Subtract 16 from both sides.

$$9x^{2}-33x=-12-16$$

Subtract 16 from -12 to get -28.

$$9x^{2}-33x=-28$$

Divide both sides by 9.

$$\frac{9x^{2}-33x}{9}=-\frac{28}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{33}{9}\right)x=-\frac{28}{9}$$

Reduce the fraction \frac{-33}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{11}{3}x=-\frac{28}{9}$$

Divide -\frac{11}{3}, the coefficient of the x term, by 2 to get -\frac{11}{6}. Then add the square of -\frac{11}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{3}x+\left(-\frac{11}{6}\right)^{2}=-\frac{28}{9}+\left(-\frac{11}{6}\right)^{2}$$

Square -\frac{11}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=-\frac{28}{9}+\frac{121}{36}$$

Add -\frac{28}{9} to \frac{121}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=\frac{1}{4}$$

Factor x^{2}-\frac{11}{3}x+\frac{121}{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}.

$$\left(x-\frac{11}{6}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{6}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{11}{6}=\frac{1}{2} x-\frac{11}{6}=-\frac{1}{2}$$

Add \frac{11}{6} to both sides of the equation.

$$x=\frac{7}{3} x=\frac{4}{3}$$