Solve x-2/3/x+frac{2{3}}=3x-12/3x^2+20x+12

$$\frac { x - \frac { 2 } { 3 } } { x + \frac { 2 } { 3 } } = \frac { 3 x - 12 } { 3 x ^ { 2 } + 20 x + 12 }$$

Solve for x

$$x = -\frac{13}{3} = -4\frac{1}{3} \approx -4.333333333$$

$$x=0$$

Steps Using Factoring

Factor the expressions that are not already factored in \frac{x-\frac{2}{3}}{x+\frac{2}{3}}.

$$\frac{\frac{1}{3}\left(3x-2\right)}{\frac{1}{3}\left(3x+2\right)}=\frac{3x-12}{3x^{2}+20x+12}$$

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$\frac{3x-2}{\left(\frac{1}{3}\right)^{0}\left(3x+2\right)}=\frac{3x-12}{3x^{2}+20x+12}$$

Calculate \frac{1}{3} to the power of 0 and get 1.

$$\frac{3x-2}{1\left(3x+2\right)}=\frac{3x-12}{3x^{2}+20x+12}$$

Use the distributive property to multiply 1 by 3x+2.

$$\frac{3x-2}{3x+2}=\frac{3x-12}{3x^{2}+20x+12}$$

Subtract \frac{3x-12}{3x^{2}+20x+12} from both sides.

$$\frac{3x-2}{3x+2}-\frac{3x-12}{3x^{2}+20x+12}=0$$

Factor 3x^{2}+20x+12.

$$\frac{3x-2}{3x+2}-\frac{3x-12}{\left(x+6\right)\left(3x+2\right)}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3x+2 and \left(x+6\right)\left(3x+2\right) is \left(x+6\right)\left(3x+2\right). Multiply \frac{3x-2}{3x+2} times \frac{x+6}{x+6}.

$$\frac{\left(3x-2\right)\left(x+6\right)}{\left(x+6\right)\left(3x+2\right)}-\frac{3x-12}{\left(x+6\right)\left(3x+2\right)}=0$$

Since \frac{\left(3x-2\right)\left(x+6\right)}{\left(x+6\right)\left(3x+2\right)} and \frac{3x-12}{\left(x+6\right)\left(3x+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(3x-2\right)\left(x+6\right)-\left(3x-12\right)}{\left(x+6\right)\left(3x+2\right)}=0$$

Do the multiplications in \left(3x-2\right)\left(x+6\right)-\left(3x-12\right).

$$\frac{3x^{2}+18x-2x-12-3x+12}{\left(x+6\right)\left(3x+2\right)}=0$$

Combine like terms in 3x^{2}+18x-2x-12-3x+12.

$$\frac{3x^{2}+13x}{\left(x+6\right)\left(3x+2\right)}=0$$

Variable x cannot be equal to any of the values -6,-\frac{2}{3} since division by zero is not defined. Multiply both sides of the equation by \left(x+6\right)\left(3x+2\right).

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

Factor out x.

$$x\left(3x+13\right)=0$$

To find equation solutions, solve x=0 and 3x+13=0.

$$x=0 x=-\frac{13}{3}$$

Steps Using the Quadratic Formula

Factor the expressions that are not already factored in \frac{x-\frac{2}{3}}{x+\frac{2}{3}}.

$$\frac{\frac{1}{3}\left(3x-2\right)}{\frac{1}{3}\left(3x+2\right)}=\frac{3x-12}{3x^{2}+20x+12}$$

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$\frac{3x-2}{\left(\frac{1}{3}\right)^{0}\left(3x+2\right)}=\frac{3x-12}{3x^{2}+20x+12}$$

Calculate \frac{1}{3} to the power of 0 and get 1.

$$\frac{3x-2}{1\left(3x+2\right)}=\frac{3x-12}{3x^{2}+20x+12}$$

Use the distributive property to multiply 1 by 3x+2.

$$\frac{3x-2}{3x+2}=\frac{3x-12}{3x^{2}+20x+12}$$

Subtract \frac{3x-12}{3x^{2}+20x+12} from both sides.

$$\frac{3x-2}{3x+2}-\frac{3x-12}{3x^{2}+20x+12}=0$$

Factor 3x^{2}+20x+12.

$$\frac{3x-2}{3x+2}-\frac{3x-12}{\left(x+6\right)\left(3x+2\right)}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3x+2 and \left(x+6\right)\left(3x+2\right) is \left(x+6\right)\left(3x+2\right). Multiply \frac{3x-2}{3x+2} times \frac{x+6}{x+6}.

$$\frac{\left(3x-2\right)\left(x+6\right)}{\left(x+6\right)\left(3x+2\right)}-\frac{3x-12}{\left(x+6\right)\left(3x+2\right)}=0$$

Since \frac{\left(3x-2\right)\left(x+6\right)}{\left(x+6\right)\left(3x+2\right)} and \frac{3x-12}{\left(x+6\right)\left(3x+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(3x-2\right)\left(x+6\right)-\left(3x-12\right)}{\left(x+6\right)\left(3x+2\right)}=0$$

Do the multiplications in \left(3x-2\right)\left(x+6\right)-\left(3x-12\right).

$$\frac{3x^{2}+18x-2x-12-3x+12}{\left(x+6\right)\left(3x+2\right)}=0$$

Combine like terms in 3x^{2}+18x-2x-12-3x+12.

$$\frac{3x^{2}+13x}{\left(x+6\right)\left(3x+2\right)}=0$$

Variable x cannot be equal to any of the values -6,-\frac{2}{3} since division by zero is not defined. Multiply both sides of the equation by \left(x+6\right)\left(3x+2\right).

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 13 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-13±\sqrt{13^{2}}}{2\times 3}$$

Take the square root of 13^{2}.

$$x=\frac{-13±13}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-13±13}{6}$$

Now solve the equation x=\frac{-13±13}{6} when ± is plus. Add -13 to 13.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

Now solve the equation x=\frac{-13±13}{6} when ± is minus. Subtract 13 from -13.

$$x=-\frac{26}{6}$$

Reduce the fraction \frac{-26}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{13}{3}$$

The equation is now solved.

$$x=0 x=-\frac{13}{3}$$

Steps for Completing the Square

Factor the expressions that are not already factored in \frac{x-\frac{2}{3}}{x+\frac{2}{3}}.

$$\frac{\frac{1}{3}\left(3x-2\right)}{\frac{1}{3}\left(3x+2\right)}=\frac{3x-12}{3x^{2}+20x+12}$$

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$\frac{3x-2}{\left(\frac{1}{3}\right)^{0}\left(3x+2\right)}=\frac{3x-12}{3x^{2}+20x+12}$$

Calculate \frac{1}{3} to the power of 0 and get 1.

$$\frac{3x-2}{1\left(3x+2\right)}=\frac{3x-12}{3x^{2}+20x+12}$$

Use the distributive property to multiply 1 by 3x+2.

$$\frac{3x-2}{3x+2}=\frac{3x-12}{3x^{2}+20x+12}$$

Subtract \frac{3x-12}{3x^{2}+20x+12} from both sides.

$$\frac{3x-2}{3x+2}-\frac{3x-12}{3x^{2}+20x+12}=0$$

Factor 3x^{2}+20x+12.

$$\frac{3x-2}{3x+2}-\frac{3x-12}{\left(x+6\right)\left(3x+2\right)}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3x+2 and \left(x+6\right)\left(3x+2\right) is \left(x+6\right)\left(3x+2\right). Multiply \frac{3x-2}{3x+2} times \frac{x+6}{x+6}.

$$\frac{\left(3x-2\right)\left(x+6\right)}{\left(x+6\right)\left(3x+2\right)}-\frac{3x-12}{\left(x+6\right)\left(3x+2\right)}=0$$

Since \frac{\left(3x-2\right)\left(x+6\right)}{\left(x+6\right)\left(3x+2\right)} and \frac{3x-12}{\left(x+6\right)\left(3x+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(3x-2\right)\left(x+6\right)-\left(3x-12\right)}{\left(x+6\right)\left(3x+2\right)}=0$$

Do the multiplications in \left(3x-2\right)\left(x+6\right)-\left(3x-12\right).

$$\frac{3x^{2}+18x-2x-12-3x+12}{\left(x+6\right)\left(3x+2\right)}=0$$

Combine like terms in 3x^{2}+18x-2x-12-3x+12.

$$\frac{3x^{2}+13x}{\left(x+6\right)\left(3x+2\right)}=0$$

Variable x cannot be equal to any of the values -6,-\frac{2}{3} since division by zero is not defined. Multiply both sides of the equation by \left(x+6\right)\left(3x+2\right).

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

Divide both sides by 3.

$$\frac{3x^{2}+13x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{13}{3}x=\frac{0}{3}$$

Divide 0 by 3.

$$x^{2}+\frac{13}{3}x=0$$

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.

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

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

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

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}.

$$\left(x+\frac{13}{6}\right)^{2}=\frac{169}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{6}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Simplify.

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

Subtract \frac{13}{6} from both sides of the equation.

$$x=0 x=-\frac{13}{3}$$