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

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

Solve for x

$$x=1$$

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

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}.

$$9x^{2}+12x+4-5\left(3x+2\right)=0$$

Use the distributive property to multiply -5 by 3x+2.

$$9x^{2}+12x+4-15x-10=0$$

Combine 12x and -15x to get -3x.

$$9x^{2}-3x+4-10=0$$

Subtract 10 from 4 to get -6.

$$9x^{2}-3x-6=0$$

Divide both sides by 3.

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

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.

$$a+b=-1 ab=3\left(-2\right)=-6$$

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 -6.

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

Calculate the sum for each pair.

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

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

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

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

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

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

Factor out common term x-1 by using distributive property.

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

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

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

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}.

$$9x^{2}+12x+4-5\left(3x+2\right)=0$$

Use the distributive property to multiply -5 by 3x+2.

$$9x^{2}+12x+4-15x-10=0$$

Combine 12x and -15x to get -3x.

$$9x^{2}-3x+4-10=0$$

Subtract 10 from 4 to get -6.

$$9x^{2}-3x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -3 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 9\left(-6\right)}}{2\times 9}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 9\left(-6\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-3\right)±\sqrt{9-36\left(-6\right)}}{2\times 9}$$

Multiply -36 times -6.

$$x=\frac{-\left(-3\right)±\sqrt{9+216}}{2\times 9}$$

Add 9 to 216.

$$x=\frac{-\left(-3\right)±\sqrt{225}}{2\times 9}$$

Take the square root of 225.

$$x=\frac{-\left(-3\right)±15}{2\times 9}$$

The opposite of -3 is 3.

$$x=\frac{3±15}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{3±15}{18}$$

Now solve the equation x=\frac{3±15}{18} when ± is plus. Add 3 to 15.

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

Divide 18 by 18.

$$x=1$$

Now solve the equation x=\frac{3±15}{18} when ± is minus. Subtract 15 from 3.

$$x=-\frac{12}{18}$$

Reduce the fraction \frac{-12}{18} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

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

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}.

$$9x^{2}+12x+4-5\left(3x+2\right)=0$$

Use the distributive property to multiply -5 by 3x+2.

$$9x^{2}+12x+4-15x-10=0$$

Combine 12x and -15x to get -3x.

$$9x^{2}-3x+4-10=0$$

Subtract 10 from 4 to get -6.

$$9x^{2}-3x-6=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$9x^{2}-3x=6$$

Divide both sides by 9.

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

Dividing by 9 undoes the multiplication by 9.

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

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

$$x^{2}-\frac{1}{3}x=\frac{6}{9}$$

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

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

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{2}{3}+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

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

Add \frac{2}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{25}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{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}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{25}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{5}{6} x-\frac{1}{6}=-\frac{5}{6}$$

Add \frac{1}{6} to both sides of the equation.

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