Solve (3x-4)^2-5(3x-4)+6=0

$$( 3 x - 4 ) ^ { 2 } - 5 ( 3 x - 4 ) + 6 = 0$$

Solve for x

$$x=2$$

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

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}.

$$9x^{2}-24x+16-5\left(3x-4\right)+6=0$$

Use the distributive property to multiply -5 by 3x-4.

$$9x^{2}-24x+16-15x+20+6=0$$

Combine -24x and -15x to get -39x.

$$9x^{2}-39x+16+20+6=0$$

Add 16 and 20 to get 36.

$$9x^{2}-39x+36+6=0$$

Add 36 and 6 to get 42.

$$9x^{2}-39x+42=0$$

Divide both sides by 3.

$$3x^{2}-13x+14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+14. To find a and b, set up a system to be solved.

$$a+b=-13 ab=3\times 14=42$$

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 42.

$$-1,-42 -2,-21 -3,-14 -6,-7$$

Calculate the sum for each pair.

$$-1-42=-43 -2-21=-23 -3-14=-17 -6-7=-13$$

The solution is the pair that gives sum -13.

$$a=-7 b=-6$$

Rewrite 3x^{2}-13x+14 as \left(3x^{2}-7x\right)+\left(-6x+14\right).

$$\left(3x^{2}-7x\right)+\left(-6x+14\right)$$

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

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

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

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

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

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

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}.

$$9x^{2}-24x+16-5\left(3x-4\right)+6=0$$

Use the distributive property to multiply -5 by 3x-4.

$$9x^{2}-24x+16-15x+20+6=0$$

Combine -24x and -15x to get -39x.

$$9x^{2}-39x+16+20+6=0$$

Add 16 and 20 to get 36.

$$9x^{2}-39x+36+6=0$$

Add 36 and 6 to get 42.

$$9x^{2}-39x+42=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -39 for b, and 42 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-39\right)±\sqrt{\left(-39\right)^{2}-4\times 9\times 42}}{2\times 9}$$

Square -39.

$$x=\frac{-\left(-39\right)±\sqrt{1521-4\times 9\times 42}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-39\right)±\sqrt{1521-36\times 42}}{2\times 9}$$

Multiply -36 times 42.

$$x=\frac{-\left(-39\right)±\sqrt{1521-1512}}{2\times 9}$$

Add 1521 to -1512.

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

Take the square root of 9.

$$x=\frac{-\left(-39\right)±3}{2\times 9}$$

The opposite of -39 is 39.

$$x=\frac{39±3}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{39±3}{18}$$

Now solve the equation x=\frac{39±3}{18} when ± is plus. Add 39 to 3.

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

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{39±3}{18} when ± is minus. Subtract 3 from 39.

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

Divide 36 by 18.

$$x=2$$

The equation is now solved.

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

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}.

$$9x^{2}-24x+16-5\left(3x-4\right)+6=0$$

Use the distributive property to multiply -5 by 3x-4.

$$9x^{2}-24x+16-15x+20+6=0$$

Combine -24x and -15x to get -39x.

$$9x^{2}-39x+16+20+6=0$$

Add 16 and 20 to get 36.

$$9x^{2}-39x+36+6=0$$

Add 36 and 6 to get 42.

$$9x^{2}-39x+42=0$$

Subtract 42 from both sides. Anything subtracted from zero gives its negation.

$$9x^{2}-39x=-42$$

Divide both sides by 9.

$$\frac{9x^{2}-39x}{9}=-\frac{42}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{39}{9}\right)x=-\frac{42}{9}$$

Reduce the fraction \frac{-39}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{13}{3}x=-\frac{42}{9}$$

Reduce the fraction \frac{-42}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{13}{3}x=-\frac{14}{3}$$

Divide -\frac{13}{3}, the coefficient of the x term, by 2 to get -\frac{13}{6}. Then add the square of -\frac{13}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{3}x+\left(-\frac{13}{6}\right)^{2}=-\frac{14}{3}+\left(-\frac{13}{6}\right)^{2}$$

Square -\frac{13}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{3}x+\frac{169}{36}=-\frac{14}{3}+\frac{169}{36}$$

Add -\frac{14}{3} to \frac{169}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{13}{3}x+\frac{169}{36}=\frac{1}{36}$$

Factor x^{2}-\frac{13}{3}x+\frac{169}{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}.

$$\left(x-\frac{13}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x-\frac{13}{6}=\frac{1}{6} x-\frac{13}{6}=-\frac{1}{6}$$

Add \frac{13}{6} to both sides of the equation.

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