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

$$( \frac { x - 2 } { x + 2 } ) ^ { 2 } - 4 ( \frac { x - 2 } { x + 2 } ) + 3 = 0$$

Solve for x

$$x=-4$$

Steps Using Factoring By Grouping

Variable x cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by x+2.

$$\left(x+2\right)\left(\left(\frac{x-2}{x+2}\right)^{2}-4\times \frac{x-2}{x+2}\right)+\left(x+2\right)\times 3=0$$

To raise \frac{x-2}{x+2} to a power, raise both numerator and denominator to the power and then divide.

$$\left(x+2\right)\left(\frac{\left(x-2\right)^{2}}{\left(x+2\right)^{2}}-4\times \frac{x-2}{x+2}\right)+\left(x+2\right)\times 3=0$$

Express 4\times \frac{x-2}{x+2} as a single fraction.

$$\left(x+2\right)\left(\frac{\left(x-2\right)^{2}}{\left(x+2\right)^{2}}-\frac{4\left(x-2\right)}{x+2}\right)+\left(x+2\right)\times 3=0$$

Use the distributive property to multiply 4 by x-2.

$$\left(x+2\right)\left(\frac{\left(x-2\right)^{2}}{\left(x+2\right)^{2}}-\frac{4x-8}{x+2}\right)+\left(x+2\right)\times 3=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+2\right)^{2} and x+2 is \left(x+2\right)^{2}. Multiply \frac{4x-8}{x+2} times \frac{x+2}{x+2}.

$$\left(x+2\right)\left(\frac{\left(x-2\right)^{2}}{\left(x+2\right)^{2}}-\frac{\left(4x-8\right)\left(x+2\right)}{\left(x+2\right)^{2}}\right)+\left(x+2\right)\times 3=0$$

Since \frac{\left(x-2\right)^{2}}{\left(x+2\right)^{2}} and \frac{\left(4x-8\right)\left(x+2\right)}{\left(x+2\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\left(x+2\right)\times \frac{\left(x-2\right)^{2}-\left(4x-8\right)\left(x+2\right)}{\left(x+2\right)^{2}}+\left(x+2\right)\times 3=0$$

Do the multiplications in \left(x-2\right)^{2}-\left(4x-8\right)\left(x+2\right).

$$\left(x+2\right)\times \frac{x^{2}-4x+4-4x^{2}-8x+8x+16}{\left(x+2\right)^{2}}+\left(x+2\right)\times 3=0$$

Combine like terms in x^{2}-4x+4-4x^{2}-8x+8x+16.

$$\left(x+2\right)\times \frac{-3x^{2}-4x+20}{\left(x+2\right)^{2}}+\left(x+2\right)\times 3=0$$

Express \left(x+2\right)\times \frac{-3x^{2}-4x+20}{\left(x+2\right)^{2}} as a single fraction.

$$\frac{\left(x+2\right)\left(-3x^{2}-4x+20\right)}{\left(x+2\right)^{2}}+\left(x+2\right)\times 3=0$$

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

$$\frac{\left(x+2\right)\left(-3x^{2}-4x+20\right)}{\left(x+2\right)^{2}}+3x+6=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3x+6 times \frac{\left(x+2\right)^{2}}{\left(x+2\right)^{2}}.

$$\frac{\left(x+2\right)\left(-3x^{2}-4x+20\right)}{\left(x+2\right)^{2}}+\frac{\left(3x+6\right)\left(x+2\right)^{2}}{\left(x+2\right)^{2}}=0$$

Since \frac{\left(x+2\right)\left(-3x^{2}-4x+20\right)}{\left(x+2\right)^{2}} and \frac{\left(3x+6\right)\left(x+2\right)^{2}}{\left(x+2\right)^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(x+2\right)\left(-3x^{2}-4x+20\right)+\left(3x+6\right)\left(x+2\right)^{2}}{\left(x+2\right)^{2}}=0$$

Do the multiplications in \left(x+2\right)\left(-3x^{2}-4x+20\right)+\left(3x+6\right)\left(x+2\right)^{2}.

$$\frac{-3x^{3}-4x^{2}+20x-6x^{2}-8x+40+3x^{3}+12x^{2}+12x+6x^{2}+24x+24}{\left(x+2\right)^{2}}=0$$

Combine like terms in -3x^{3}-4x^{2}+20x-6x^{2}-8x+40+3x^{3}+12x^{2}+12x+6x^{2}+24x+24.

$$\frac{8x^{2}+48x+64}{\left(x+2\right)^{2}}=0$$

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

$$\frac{8x^{2}+48x+64}{x^{2}+4x+4}=0$$

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

$$8x^{2}+48x+64=0$$

Divide both sides by 8.

$$x^{2}+6x+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\times 8=8$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 8.

$$1,8 2,4$$

Calculate the sum for each pair.

$$1+8=9 2+4=6$$

The solution is the pair that gives sum 6.

$$a=2 b=4$$

Rewrite x^{2}+6x+8 as \left(x^{2}+2x\right)+\left(4x+8\right).

$$\left(x^{2}+2x\right)+\left(4x+8\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(x+2\right)+4\left(x+2\right)$$

Factor out common term x+2 by using distributive property.

$$\left(x+2\right)\left(x+4\right)$$

To find equation solutions, solve x+2=0 and x+4=0.

$$x=-2 x=-4$$

Variable x cannot be equal to -2.

$$x=-4$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by x+2.

$$\left(x+2\right)\left(\left(\frac{x-2}{x+2}\right)^{2}-4\times \frac{x-2}{x+2}\right)+\left(x+2\right)\times 3=0$$

To raise \frac{x-2}{x+2} to a power, raise both numerator and denominator to the power and then divide.

$$\left(x+2\right)\left(\frac{\left(x-2\right)^{2}}{\left(x+2\right)^{2}}-4\times \frac{x-2}{x+2}\right)+\left(x+2\right)\times 3=0$$

Express 4\times \frac{x-2}{x+2} as a single fraction.

$$\left(x+2\right)\left(\frac{\left(x-2\right)^{2}}{\left(x+2\right)^{2}}-\frac{4\left(x-2\right)}{x+2}\right)+\left(x+2\right)\times 3=0$$

Use the distributive property to multiply 4 by x-2.

$$\left(x+2\right)\left(\frac{\left(x-2\right)^{2}}{\left(x+2\right)^{2}}-\frac{4x-8}{x+2}\right)+\left(x+2\right)\times 3=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+2\right)^{2} and x+2 is \left(x+2\right)^{2}. Multiply \frac{4x-8}{x+2} times \frac{x+2}{x+2}.

$$\left(x+2\right)\left(\frac{\left(x-2\right)^{2}}{\left(x+2\right)^{2}}-\frac{\left(4x-8\right)\left(x+2\right)}{\left(x+2\right)^{2}}\right)+\left(x+2\right)\times 3=0$$

Since \frac{\left(x-2\right)^{2}}{\left(x+2\right)^{2}} and \frac{\left(4x-8\right)\left(x+2\right)}{\left(x+2\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\left(x+2\right)\times \frac{\left(x-2\right)^{2}-\left(4x-8\right)\left(x+2\right)}{\left(x+2\right)^{2}}+\left(x+2\right)\times 3=0$$

Do the multiplications in \left(x-2\right)^{2}-\left(4x-8\right)\left(x+2\right).

$$\left(x+2\right)\times \frac{x^{2}-4x+4-4x^{2}-8x+8x+16}{\left(x+2\right)^{2}}+\left(x+2\right)\times 3=0$$

Combine like terms in x^{2}-4x+4-4x^{2}-8x+8x+16.

$$\left(x+2\right)\times \frac{-3x^{2}-4x+20}{\left(x+2\right)^{2}}+\left(x+2\right)\times 3=0$$

Express \left(x+2\right)\times \frac{-3x^{2}-4x+20}{\left(x+2\right)^{2}} as a single fraction.

$$\frac{\left(x+2\right)\left(-3x^{2}-4x+20\right)}{\left(x+2\right)^{2}}+\left(x+2\right)\times 3=0$$

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

$$\frac{\left(x+2\right)\left(-3x^{2}-4x+20\right)}{\left(x+2\right)^{2}}+3x+6=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3x+6 times \frac{\left(x+2\right)^{2}}{\left(x+2\right)^{2}}.

$$\frac{\left(x+2\right)\left(-3x^{2}-4x+20\right)}{\left(x+2\right)^{2}}+\frac{\left(3x+6\right)\left(x+2\right)^{2}}{\left(x+2\right)^{2}}=0$$

Since \frac{\left(x+2\right)\left(-3x^{2}-4x+20\right)}{\left(x+2\right)^{2}} and \frac{\left(3x+6\right)\left(x+2\right)^{2}}{\left(x+2\right)^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(x+2\right)\left(-3x^{2}-4x+20\right)+\left(3x+6\right)\left(x+2\right)^{2}}{\left(x+2\right)^{2}}=0$$

Do the multiplications in \left(x+2\right)\left(-3x^{2}-4x+20\right)+\left(3x+6\right)\left(x+2\right)^{2}.

$$\frac{-3x^{3}-4x^{2}+20x-6x^{2}-8x+40+3x^{3}+12x^{2}+12x+6x^{2}+24x+24}{\left(x+2\right)^{2}}=0$$

Combine like terms in -3x^{3}-4x^{2}+20x-6x^{2}-8x+40+3x^{3}+12x^{2}+12x+6x^{2}+24x+24.

$$\frac{8x^{2}+48x+64}{\left(x+2\right)^{2}}=0$$

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

$$\frac{8x^{2}+48x+64}{x^{2}+4x+4}=0$$

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

$$8x^{2}+48x+64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 48 for b, and 64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-48±\sqrt{48^{2}-4\times 8\times 64}}{2\times 8}$$

Square 48.

$$x=\frac{-48±\sqrt{2304-4\times 8\times 64}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-48±\sqrt{2304-32\times 64}}{2\times 8}$$

Multiply -32 times 64.

$$x=\frac{-48±\sqrt{2304-2048}}{2\times 8}$$

Add 2304 to -2048.

$$x=\frac{-48±\sqrt{256}}{2\times 8}$$

Take the square root of 256.

$$x=\frac{-48±16}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-48±16}{16}$$

Now solve the equation x=\frac{-48±16}{16} when ± is plus. Add -48 to 16.

$$x=-\frac{32}{16}$$

Divide -32 by 16.

$$x=-2$$

Now solve the equation x=\frac{-48±16}{16} when ± is minus. Subtract 16 from -48.

$$x=-\frac{64}{16}$$

Divide -64 by 16.

$$x=-4$$

The equation is now solved.

$$x=-2 x=-4$$

Variable x cannot be equal to -2.

$$x=-4$$

Steps for Completing the Square

Variable x cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by x+2.

$$\left(x+2\right)\left(\left(\frac{x-2}{x+2}\right)^{2}-4\times \frac{x-2}{x+2}\right)+\left(x+2\right)\times 3=0$$

To raise \frac{x-2}{x+2} to a power, raise both numerator and denominator to the power and then divide.

$$\left(x+2\right)\left(\frac{\left(x-2\right)^{2}}{\left(x+2\right)^{2}}-4\times \frac{x-2}{x+2}\right)+\left(x+2\right)\times 3=0$$

Express 4\times \frac{x-2}{x+2} as a single fraction.

$$\left(x+2\right)\left(\frac{\left(x-2\right)^{2}}{\left(x+2\right)^{2}}-\frac{4\left(x-2\right)}{x+2}\right)+\left(x+2\right)\times 3=0$$

Use the distributive property to multiply 4 by x-2.

$$\left(x+2\right)\left(\frac{\left(x-2\right)^{2}}{\left(x+2\right)^{2}}-\frac{4x-8}{x+2}\right)+\left(x+2\right)\times 3=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+2\right)^{2} and x+2 is \left(x+2\right)^{2}. Multiply \frac{4x-8}{x+2} times \frac{x+2}{x+2}.

$$\left(x+2\right)\left(\frac{\left(x-2\right)^{2}}{\left(x+2\right)^{2}}-\frac{\left(4x-8\right)\left(x+2\right)}{\left(x+2\right)^{2}}\right)+\left(x+2\right)\times 3=0$$

Since \frac{\left(x-2\right)^{2}}{\left(x+2\right)^{2}} and \frac{\left(4x-8\right)\left(x+2\right)}{\left(x+2\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\left(x+2\right)\times \frac{\left(x-2\right)^{2}-\left(4x-8\right)\left(x+2\right)}{\left(x+2\right)^{2}}+\left(x+2\right)\times 3=0$$

Do the multiplications in \left(x-2\right)^{2}-\left(4x-8\right)\left(x+2\right).

$$\left(x+2\right)\times \frac{x^{2}-4x+4-4x^{2}-8x+8x+16}{\left(x+2\right)^{2}}+\left(x+2\right)\times 3=0$$

Combine like terms in x^{2}-4x+4-4x^{2}-8x+8x+16.

$$\left(x+2\right)\times \frac{-3x^{2}-4x+20}{\left(x+2\right)^{2}}+\left(x+2\right)\times 3=0$$

Express \left(x+2\right)\times \frac{-3x^{2}-4x+20}{\left(x+2\right)^{2}} as a single fraction.

$$\frac{\left(x+2\right)\left(-3x^{2}-4x+20\right)}{\left(x+2\right)^{2}}+\left(x+2\right)\times 3=0$$

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

$$\frac{\left(x+2\right)\left(-3x^{2}-4x+20\right)}{\left(x+2\right)^{2}}+3x+6=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3x+6 times \frac{\left(x+2\right)^{2}}{\left(x+2\right)^{2}}.

$$\frac{\left(x+2\right)\left(-3x^{2}-4x+20\right)}{\left(x+2\right)^{2}}+\frac{\left(3x+6\right)\left(x+2\right)^{2}}{\left(x+2\right)^{2}}=0$$

Since \frac{\left(x+2\right)\left(-3x^{2}-4x+20\right)}{\left(x+2\right)^{2}} and \frac{\left(3x+6\right)\left(x+2\right)^{2}}{\left(x+2\right)^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(x+2\right)\left(-3x^{2}-4x+20\right)+\left(3x+6\right)\left(x+2\right)^{2}}{\left(x+2\right)^{2}}=0$$

Do the multiplications in \left(x+2\right)\left(-3x^{2}-4x+20\right)+\left(3x+6\right)\left(x+2\right)^{2}.

$$\frac{-3x^{3}-4x^{2}+20x-6x^{2}-8x+40+3x^{3}+12x^{2}+12x+6x^{2}+24x+24}{\left(x+2\right)^{2}}=0$$

Combine like terms in -3x^{3}-4x^{2}+20x-6x^{2}-8x+40+3x^{3}+12x^{2}+12x+6x^{2}+24x+24.

$$\frac{8x^{2}+48x+64}{\left(x+2\right)^{2}}=0$$

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

$$\frac{8x^{2}+48x+64}{x^{2}+4x+4}=0$$

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

$$8x^{2}+48x+64=0$$

Subtract 64 from both sides. Anything subtracted from zero gives its negation.

$$8x^{2}+48x=-64$$

Divide both sides by 8.

$$\frac{8x^{2}+48x}{8}=-\frac{64}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{48}{8}x=-\frac{64}{8}$$

Divide 48 by 8.

$$x^{2}+6x=-\frac{64}{8}$$

Divide -64 by 8.

$$x^{2}+6x=-8$$

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

$$x^{2}+6x+3^{2}=-8+3^{2}$$

Square 3.

$$x^{2}+6x+9=-8+9$$

Add -8 to 9.

$$x^{2}+6x+9=1$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{1}$$

Simplify.

$$x+3=1 x+3=-1$$

Subtract 3 from both sides of the equation.

$$x=-2 x=-4$$

Variable x cannot be equal to -2.

$$x=-4$$