Solve x=2x+20/2x-8-x

$$x = \frac { 2 x + 20 } { 2 x - 8 } - x$$

Solve for x

$$x = \frac{\sqrt{161} + 9}{4} \approx 5.422144385$$

$$x=\frac{9-\sqrt{161}}{4}\approx -0.922144385$$

Steps Using the Quadratic Formula

Factor the expressions that are not already factored in \frac{2x+20}{2x-8}.

$$x=\frac{2\left(x+10\right)}{2\left(x-4\right)}-x$$

Cancel out 2 in both numerator and denominator.

$$x=\frac{x+10}{x-4}-x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x-4}{x-4}.

$$x=\frac{x+10}{x-4}-\frac{x\left(x-4\right)}{x-4}$$

Since \frac{x+10}{x-4} and \frac{x\left(x-4\right)}{x-4} have the same denominator, subtract them by subtracting their numerators.

$$x=\frac{x+10-x\left(x-4\right)}{x-4}$$

Do the multiplications in x+10-x\left(x-4\right).

$$x=\frac{x+10-x^{2}+4x}{x-4}$$

Combine like terms in x+10-x^{2}+4x.

$$x=\frac{5x+10-x^{2}}{x-4}$$

Subtract \frac{5x+10-x^{2}}{x-4} from both sides.

$$x-\frac{5x+10-x^{2}}{x-4}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x-4}{x-4}.

$$\frac{x\left(x-4\right)}{x-4}-\frac{5x+10-x^{2}}{x-4}=0$$

Since \frac{x\left(x-4\right)}{x-4} and \frac{5x+10-x^{2}}{x-4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x-4\right)-\left(5x+10-x^{2}\right)}{x-4}=0$$

Do the multiplications in x\left(x-4\right)-\left(5x+10-x^{2}\right).

$$\frac{x^{2}-4x-5x-10+x^{2}}{x-4}=0$$

Combine like terms in x^{2}-4x-5x-10+x^{2}.

$$\frac{2x^{2}-9x-10}{x-4}=0$$

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by x-4.

$$2x^{2}-9x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -9 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 2\left(-10\right)}}{2\times 2}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 2\left(-10\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-9\right)±\sqrt{81-8\left(-10\right)}}{2\times 2}$$

Multiply -8 times -10.

$$x=\frac{-\left(-9\right)±\sqrt{81+80}}{2\times 2}$$

Add 81 to 80.

$$x=\frac{-\left(-9\right)±\sqrt{161}}{2\times 2}$$

The opposite of -9 is 9.

$$x=\frac{9±\sqrt{161}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{9±\sqrt{161}}{4}$$

Now solve the equation x=\frac{9±\sqrt{161}}{4} when ± is plus. Add 9 to \sqrt{161}.

$$x=\frac{\sqrt{161}+9}{4}$$

Now solve the equation x=\frac{9±\sqrt{161}}{4} when ± is minus. Subtract \sqrt{161} from 9.

$$x=\frac{9-\sqrt{161}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{161}+9}{4} x=\frac{9-\sqrt{161}}{4}$$

Steps for Completing the Square

Factor the expressions that are not already factored in \frac{2x+20}{2x-8}.

$$x=\frac{2\left(x+10\right)}{2\left(x-4\right)}-x$$

Cancel out 2 in both numerator and denominator.

$$x=\frac{x+10}{x-4}-x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x-4}{x-4}.

$$x=\frac{x+10}{x-4}-\frac{x\left(x-4\right)}{x-4}$$

Since \frac{x+10}{x-4} and \frac{x\left(x-4\right)}{x-4} have the same denominator, subtract them by subtracting their numerators.

$$x=\frac{x+10-x\left(x-4\right)}{x-4}$$

Do the multiplications in x+10-x\left(x-4\right).

$$x=\frac{x+10-x^{2}+4x}{x-4}$$

Combine like terms in x+10-x^{2}+4x.

$$x=\frac{5x+10-x^{2}}{x-4}$$

Subtract \frac{5x+10-x^{2}}{x-4} from both sides.

$$x-\frac{5x+10-x^{2}}{x-4}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x-4}{x-4}.

$$\frac{x\left(x-4\right)}{x-4}-\frac{5x+10-x^{2}}{x-4}=0$$

Since \frac{x\left(x-4\right)}{x-4} and \frac{5x+10-x^{2}}{x-4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x-4\right)-\left(5x+10-x^{2}\right)}{x-4}=0$$

Do the multiplications in x\left(x-4\right)-\left(5x+10-x^{2}\right).

$$\frac{x^{2}-4x-5x-10+x^{2}}{x-4}=0$$

Combine like terms in x^{2}-4x-5x-10+x^{2}.

$$\frac{2x^{2}-9x-10}{x-4}=0$$

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by x-4.

$$2x^{2}-9x-10=0$$

Add 10 to both sides. Anything plus zero gives itself.

$$2x^{2}-9x=10$$

Divide both sides by 2.

$$\frac{2x^{2}-9x}{2}=\frac{10}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{9}{2}x=\frac{10}{2}$$

Divide 10 by 2.

$$x^{2}-\frac{9}{2}x=5$$

Divide -\frac{9}{2}, the coefficient of the x term, by 2 to get -\frac{9}{4}. Then add the square of -\frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{2}x+\left(-\frac{9}{4}\right)^{2}=5+\left(-\frac{9}{4}\right)^{2}$$

Square -\frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=5+\frac{81}{16}$$

Add 5 to \frac{81}{16}.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{161}{16}$$

Factor x^{2}-\frac{9}{2}x+\frac{81}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{4}\right)^{2}=\frac{161}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{4}\right)^{2}}=\sqrt{\frac{161}{16}}$$

Simplify.

$$x-\frac{9}{4}=\frac{\sqrt{161}}{4} x-\frac{9}{4}=-\frac{\sqrt{161}}{4}$$

Add \frac{9}{4} to both sides of the equation.

$$x=\frac{\sqrt{161}+9}{4} x=\frac{9-\sqrt{161}}{4}$$