Solve (3x+1)(5x-3)=x(3x+1)

$$( 3 x + 1 ) ( 5 x - 3 ) = x ( 3 x + 1 )$$

Solve for x

$$x=-\frac{1}{3}\approx -0.333333333$$

$$x=\frac{3}{4}=0.75$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 3x+1 by 5x-3 and combine like terms.

$$15x^{2}-4x-3=x\left(3x+1\right)$$

Use the distributive property to multiply x by 3x+1.

$$15x^{2}-4x-3=3x^{2}+x$$

Subtract 3x^{2} from both sides.

$$15x^{2}-4x-3-3x^{2}=x$$

Combine 15x^{2} and -3x^{2} to get 12x^{2}.

$$12x^{2}-4x-3=x$$

Subtract x from both sides.

$$12x^{2}-4x-3-x=0$$

Combine -4x and -x to get -5x.

$$12x^{2}-5x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 12x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-5 ab=12\left(-3\right)=-36$$

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 -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -5.

$$a=-9 b=4$$

Rewrite 12x^{2}-5x-3 as \left(12x^{2}-9x\right)+\left(4x-3\right).

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

Factor out 3x in 12x^{2}-9x.

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

Factor out common term 4x-3 by using distributive property.

$$\left(4x-3\right)\left(3x+1\right)$$

To find equation solutions, solve 4x-3=0 and 3x+1=0.

$$x=\frac{3}{4} x=-\frac{1}{3}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x+1 by 5x-3 and combine like terms.

$$15x^{2}-4x-3=x\left(3x+1\right)$$

Use the distributive property to multiply x by 3x+1.

$$15x^{2}-4x-3=3x^{2}+x$$

Subtract 3x^{2} from both sides.

$$15x^{2}-4x-3-3x^{2}=x$$

Combine 15x^{2} and -3x^{2} to get 12x^{2}.

$$12x^{2}-4x-3=x$$

Subtract x from both sides.

$$12x^{2}-4x-3-x=0$$

Combine -4x and -x to get -5x.

$$12x^{2}-5x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -5 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 12\left(-3\right)}}{2\times 12}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 12\left(-3\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-5\right)±\sqrt{25-48\left(-3\right)}}{2\times 12}$$

Multiply -48 times -3.

$$x=\frac{-\left(-5\right)±\sqrt{25+144}}{2\times 12}$$

Add 25 to 144.

$$x=\frac{-\left(-5\right)±\sqrt{169}}{2\times 12}$$

Take the square root of 169.

$$x=\frac{-\left(-5\right)±13}{2\times 12}$$

The opposite of -5 is 5.

$$x=\frac{5±13}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{5±13}{24}$$

Now solve the equation x=\frac{5±13}{24} when ± is plus. Add 5 to 13.

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

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

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

Now solve the equation x=\frac{5±13}{24} when ± is minus. Subtract 13 from 5.

$$x=-\frac{8}{24}$$

Reduce the fraction \frac{-8}{24} to lowest terms by extracting and canceling out 8.

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

The equation is now solved.

$$x=\frac{3}{4} x=-\frac{1}{3}$$

Steps for Completing the Square

Use the distributive property to multiply 3x+1 by 5x-3 and combine like terms.

$$15x^{2}-4x-3=x\left(3x+1\right)$$

Use the distributive property to multiply x by 3x+1.

$$15x^{2}-4x-3=3x^{2}+x$$

Subtract 3x^{2} from both sides.

$$15x^{2}-4x-3-3x^{2}=x$$

Combine 15x^{2} and -3x^{2} to get 12x^{2}.

$$12x^{2}-4x-3=x$$

Subtract x from both sides.

$$12x^{2}-4x-3-x=0$$

Combine -4x and -x to get -5x.

$$12x^{2}-5x-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$12x^{2}-5x=3$$

Divide both sides by 12.

$$\frac{12x^{2}-5x}{12}=\frac{3}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}-\frac{5}{12}x=\frac{3}{12}$$

Reduce the fraction \frac{3}{12} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{5}{12}x=\frac{1}{4}$$

Divide -\frac{5}{12}, the coefficient of the x term, by 2 to get -\frac{5}{24}. Then add the square of -\frac{5}{24} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{12}x+\left(-\frac{5}{24}\right)^{2}=\frac{1}{4}+\left(-\frac{5}{24}\right)^{2}$$

Square -\frac{5}{24} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{12}x+\frac{25}{576}=\frac{1}{4}+\frac{25}{576}$$

Add \frac{1}{4} to \frac{25}{576} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{12}x+\frac{25}{576}=\frac{169}{576}$$

Factor x^{2}-\frac{5}{12}x+\frac{25}{576}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{24}\right)^{2}=\frac{169}{576}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{24}\right)^{2}}=\sqrt{\frac{169}{576}}$$

Simplify.

$$x-\frac{5}{24}=\frac{13}{24} x-\frac{5}{24}=-\frac{13}{24}$$

Add \frac{5}{24} to both sides of the equation.

$$x=\frac{3}{4} x=-\frac{1}{3}$$