Solve 10(x-2)+10/x-2=29

$$10(x-2)+ \frac{ 10 }{ x-2 } =29$$

Solve for x

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

$$x = \frac{12}{5} = 2\frac{2}{5} = 2.4$$

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.

$$10\left(x-2\right)\left(x-2\right)+10=29\left(x-2\right)$$

Multiply x-2 and x-2 to get \left(x-2\right)^{2}.

$$10\left(x-2\right)^{2}+10=29\left(x-2\right)$$

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

$$10\left(x^{2}-4x+4\right)+10=29\left(x-2\right)$$

Use the distributive property to multiply 10 by x^{2}-4x+4.

$$10x^{2}-40x+40+10=29\left(x-2\right)$$

Add 40 and 10 to get 50.

$$10x^{2}-40x+50=29\left(x-2\right)$$

Use the distributive property to multiply 29 by x-2.

$$10x^{2}-40x+50=29x-58$$

Subtract 29x from both sides.

$$10x^{2}-40x+50-29x=-58$$

Combine -40x and -29x to get -69x.

$$10x^{2}-69x+50=-58$$

Add 58 to both sides.

$$10x^{2}-69x+50+58=0$$

Add 50 and 58 to get 108.

$$10x^{2}-69x+108=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -69 for b, and 108 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-69\right)±\sqrt{\left(-69\right)^{2}-4\times 10\times 108}}{2\times 10}$$

Square -69.

$$x=\frac{-\left(-69\right)±\sqrt{4761-4\times 10\times 108}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-69\right)±\sqrt{4761-40\times 108}}{2\times 10}$$

Multiply -40 times 108.

$$x=\frac{-\left(-69\right)±\sqrt{4761-4320}}{2\times 10}$$

Add 4761 to -4320.

$$x=\frac{-\left(-69\right)±\sqrt{441}}{2\times 10}$$

Take the square root of 441.

$$x=\frac{-\left(-69\right)±21}{2\times 10}$$

The opposite of -69 is 69.

$$x=\frac{69±21}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{69±21}{20}$$

Now solve the equation x=\frac{69±21}{20} when ± is plus. Add 69 to 21.

$$x=\frac{90}{20}$$

Reduce the fraction \frac{90}{20} to lowest terms by extracting and canceling out 10.

$$x=\frac{9}{2}$$

Now solve the equation x=\frac{69±21}{20} when ± is minus. Subtract 21 from 69.

$$x=\frac{48}{20}$$

Reduce the fraction \frac{48}{20} to lowest terms by extracting and canceling out 4.

$$x=\frac{12}{5}$$

The equation is now solved.

$$x=\frac{9}{2} x=\frac{12}{5}$$

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.

$$10\left(x-2\right)\left(x-2\right)+10=29\left(x-2\right)$$

Multiply x-2 and x-2 to get \left(x-2\right)^{2}.

$$10\left(x-2\right)^{2}+10=29\left(x-2\right)$$

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

$$10\left(x^{2}-4x+4\right)+10=29\left(x-2\right)$$

Use the distributive property to multiply 10 by x^{2}-4x+4.

$$10x^{2}-40x+40+10=29\left(x-2\right)$$

Add 40 and 10 to get 50.

$$10x^{2}-40x+50=29\left(x-2\right)$$

Use the distributive property to multiply 29 by x-2.

$$10x^{2}-40x+50=29x-58$$

Subtract 29x from both sides.

$$10x^{2}-40x+50-29x=-58$$

Combine -40x and -29x to get -69x.

$$10x^{2}-69x+50=-58$$

Subtract 50 from both sides.

$$10x^{2}-69x=-58-50$$

Subtract 50 from -58 to get -108.

$$10x^{2}-69x=-108$$

Divide both sides by 10.

$$\frac{10x^{2}-69x}{10}=-\frac{108}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}-\frac{69}{10}x=-\frac{108}{10}$$

Reduce the fraction \frac{-108}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{69}{10}x=-\frac{54}{5}$$

Divide -\frac{69}{10}, the coefficient of the x term, by 2 to get -\frac{69}{20}. Then add the square of -\frac{69}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{69}{10}x+\left(-\frac{69}{20}\right)^{2}=-\frac{54}{5}+\left(-\frac{69}{20}\right)^{2}$$

Square -\frac{69}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{69}{10}x+\frac{4761}{400}=-\frac{54}{5}+\frac{4761}{400}$$

Add -\frac{54}{5} to \frac{4761}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{69}{10}x+\frac{4761}{400}=\frac{441}{400}$$

Factor x^{2}-\frac{69}{10}x+\frac{4761}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{69}{20}\right)^{2}=\frac{441}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{69}{20}\right)^{2}}=\sqrt{\frac{441}{400}}$$

Simplify.

$$x-\frac{69}{20}=\frac{21}{20} x-\frac{69}{20}=-\frac{21}{20}$$

Add \frac{69}{20} to both sides of the equation.

$$x=\frac{9}{2} x=\frac{12}{5}$$