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

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

Solve for x

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

$$x=1$$

Steps Using Factoring By Grouping

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

$$15x^{2}+10x=\left(3x+2\right)^{2}$$

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

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

Subtract 9x^{2} from both sides.

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

Combine 15x^{2} and -9x^{2} to get 6x^{2}.

$$6x^{2}+10x=12x+4$$

Subtract 12x from both sides.

$$6x^{2}+10x-12x=4$$

Combine 10x and -12x to get -2x.

$$6x^{2}-2x=4$$

Subtract 4 from both sides.

$$6x^{2}-2x-4=0$$

Divide both sides by 2.

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

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

$$15x^{2}+10x=\left(3x+2\right)^{2}$$

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

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

Subtract 9x^{2} from both sides.

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

Combine 15x^{2} and -9x^{2} to get 6x^{2}.

$$6x^{2}+10x=12x+4$$

Subtract 12x from both sides.

$$6x^{2}+10x-12x=4$$

Combine 10x and -12x to get -2x.

$$6x^{2}-2x=4$$

Subtract 4 from both sides.

$$6x^{2}-2x-4=0$$

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

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

Square -2.

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

Multiply -4 times 6.

$$x=\frac{-\left(-2\right)±\sqrt{4-24\left(-4\right)}}{2\times 6}$$

Multiply -24 times -4.

$$x=\frac{-\left(-2\right)±\sqrt{4+96}}{2\times 6}$$

Add 4 to 96.

$$x=\frac{-\left(-2\right)±\sqrt{100}}{2\times 6}$$

Take the square root of 100.

$$x=\frac{-\left(-2\right)±10}{2\times 6}$$

The opposite of -2 is 2.

$$x=\frac{2±10}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{2±10}{12}$$

Now solve the equation x=\frac{2±10}{12} when ± is plus. Add 2 to 10.

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

Divide 12 by 12.

$$x=1$$

Now solve the equation x=\frac{2±10}{12} when ± is minus. Subtract 10 from 2.

$$x=-\frac{8}{12}$$

Reduce the fraction \frac{-8}{12} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$15x^{2}+10x=\left(3x+2\right)^{2}$$

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

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

Subtract 9x^{2} from both sides.

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

Combine 15x^{2} and -9x^{2} to get 6x^{2}.

$$6x^{2}+10x=12x+4$$

Subtract 12x from both sides.

$$6x^{2}+10x-12x=4$$

Combine 10x and -12x to get -2x.

$$6x^{2}-2x=4$$

Divide both sides by 6.

$$\frac{6x^{2}-2x}{6}=\frac{4}{6}$$

Dividing by 6 undoes the multiplication by 6.

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

Reduce the fraction \frac{-2}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{3}x=\frac{4}{6}$$

Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.

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

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.

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