Solve 2x+4/x+2-x/2-x=2x^2-x-2/x^2-4

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

Solve for x

$$x = \frac{\sqrt{105} - 7}{2} \approx 1.623475383$$

$$x=\frac{-\sqrt{105}-7}{2}\approx -8.623475383$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+2\right), the least common multiple of x+2,2-x,x^{2}-4.

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

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

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

Use the distributive property to multiply 2x-4 by x+4 and combine like terms.

$$2x^{2}+4x-16-\left(-\left(2+x\right)x\right)=2x^{2}-x-2$$

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

$$2x^{2}+4x-16-\left(-2-x\right)x=2x^{2}-x-2$$

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

$$2x^{2}+4x-16-\left(-2x-x^{2}\right)=2x^{2}-x-2$$

To find the opposite of -2x-x^{2}, find the opposite of each term.

$$2x^{2}+4x-16+2x+x^{2}=2x^{2}-x-2$$

Combine 4x and 2x to get 6x.

$$2x^{2}+6x-16+x^{2}=2x^{2}-x-2$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}+6x-16=2x^{2}-x-2$$

Subtract 2x^{2} from both sides.

$$3x^{2}+6x-16-2x^{2}=-x-2$$

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

$$x^{2}+6x-16=-x-2$$

Add x to both sides.

$$x^{2}+6x-16+x=-2$$

Combine 6x and x to get 7x.

$$x^{2}+7x-16=-2$$

Add 2 to both sides.

$$x^{2}+7x-16+2=0$$

Add -16 and 2 to get -14.

$$x^{2}+7x-14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 7 for b, and -14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-14\right)}}{2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-14\right)}}{2}$$

Multiply -4 times -14.

$$x=\frac{-7±\sqrt{49+56}}{2}$$

Add 49 to 56.

$$x=\frac{-7±\sqrt{105}}{2}$$

Now solve the equation x=\frac{-7±\sqrt{105}}{2} when ± is plus. Add -7 to \sqrt{105}.

$$x=\frac{\sqrt{105}-7}{2}$$

Now solve the equation x=\frac{-7±\sqrt{105}}{2} when ± is minus. Subtract \sqrt{105} from -7.

$$x=\frac{-\sqrt{105}-7}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{105}-7}{2} x=\frac{-\sqrt{105}-7}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+2\right), the least common multiple of x+2,2-x,x^{2}-4.

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

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

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

Use the distributive property to multiply 2x-4 by x+4 and combine like terms.

$$2x^{2}+4x-16-\left(-\left(2+x\right)x\right)=2x^{2}-x-2$$

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

$$2x^{2}+4x-16-\left(-2-x\right)x=2x^{2}-x-2$$

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

$$2x^{2}+4x-16-\left(-2x-x^{2}\right)=2x^{2}-x-2$$

To find the opposite of -2x-x^{2}, find the opposite of each term.

$$2x^{2}+4x-16+2x+x^{2}=2x^{2}-x-2$$

Combine 4x and 2x to get 6x.

$$2x^{2}+6x-16+x^{2}=2x^{2}-x-2$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}+6x-16=2x^{2}-x-2$$

Subtract 2x^{2} from both sides.

$$3x^{2}+6x-16-2x^{2}=-x-2$$

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

$$x^{2}+6x-16=-x-2$$

Add x to both sides.

$$x^{2}+6x-16+x=-2$$

Combine 6x and x to get 7x.

$$x^{2}+7x-16=-2$$

Add 16 to both sides.

$$x^{2}+7x=-2+16$$

Add -2 and 16 to get 14.

$$x^{2}+7x=14$$

Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=14+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+7x+\frac{49}{4}=14+\frac{49}{4}$$

Add 14 to \frac{49}{4}.

$$x^{2}+7x+\frac{49}{4}=\frac{105}{4}$$

Factor x^{2}+7x+\frac{49}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{105}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{105}{4}}$$

Simplify.

$$x+\frac{7}{2}=\frac{\sqrt{105}}{2} x+\frac{7}{2}=-\frac{\sqrt{105}}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$x=\frac{\sqrt{105}-7}{2} x=\frac{-\sqrt{105}-7}{2}$$