Solve 2/(3k-10)^2+5/3k-10+2=0

$$\frac { 2 } { ( 3 k - 10 ) ^ { 2 } } + \frac { 5 } { 3 k - 10 } + 2 = 0$$

Solve for k

$$k = \frac{19}{6} = 3\frac{1}{6} \approx 3.166666667$$

$$k = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

Steps Using the Quadratic Formula

Variable k cannot be equal to \frac{10}{3} since division by zero is not defined. Multiply both sides of the equation by \left(3k-10\right)^{2}, the least common multiple of \left(3k-10\right)^{2},3k-10.

$$2+\left(3k-10\right)\times 5+\left(3k-10\right)^{2}\times 2=0$$

Use the distributive property to multiply 3k-10 by 5.

$$2+15k-50+\left(3k-10\right)^{2}\times 2=0$$

Subtract 50 from 2 to get -48.

$$-48+15k+\left(3k-10\right)^{2}\times 2=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3k-10\right)^{2}.

$$-48+15k+\left(9k^{2}-60k+100\right)\times 2=0$$

Use the distributive property to multiply 9k^{2}-60k+100 by 2.

$$-48+15k+18k^{2}-120k+200=0$$

Combine 15k and -120k to get -105k.

$$-48-105k+18k^{2}+200=0$$

Add -48 and 200 to get 152.

$$152-105k+18k^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$18k^{2}-105k+152=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, -105 for b, and 152 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-105\right)±\sqrt{\left(-105\right)^{2}-4\times 18\times 152}}{2\times 18}$$

Square -105.

$$k=\frac{-\left(-105\right)±\sqrt{11025-4\times 18\times 152}}{2\times 18}$$

Multiply -4 times 18.

$$k=\frac{-\left(-105\right)±\sqrt{11025-72\times 152}}{2\times 18}$$

Multiply -72 times 152.

$$k=\frac{-\left(-105\right)±\sqrt{11025-10944}}{2\times 18}$$

Add 11025 to -10944.

$$k=\frac{-\left(-105\right)±\sqrt{81}}{2\times 18}$$

Take the square root of 81.

$$k=\frac{-\left(-105\right)±9}{2\times 18}$$

The opposite of -105 is 105.

$$k=\frac{105±9}{2\times 18}$$

Multiply 2 times 18.

$$k=\frac{105±9}{36}$$

Now solve the equation k=\frac{105±9}{36} when ± is plus. Add 105 to 9.

$$k=\frac{114}{36}$$

Reduce the fraction \frac{114}{36} to lowest terms by extracting and canceling out 6.

$$k=\frac{19}{6}$$

Now solve the equation k=\frac{105±9}{36} when ± is minus. Subtract 9 from 105.

$$k=\frac{96}{36}$$

Reduce the fraction \frac{96}{36} to lowest terms by extracting and canceling out 12.

$$k=\frac{8}{3}$$

The equation is now solved.

$$k=\frac{19}{6} k=\frac{8}{3}$$

Steps for Completing the Square

Variable k cannot be equal to \frac{10}{3} since division by zero is not defined. Multiply both sides of the equation by \left(3k-10\right)^{2}, the least common multiple of \left(3k-10\right)^{2},3k-10.

$$2+\left(3k-10\right)\times 5+\left(3k-10\right)^{2}\times 2=0$$

Use the distributive property to multiply 3k-10 by 5.

$$2+15k-50+\left(3k-10\right)^{2}\times 2=0$$

Subtract 50 from 2 to get -48.

$$-48+15k+\left(3k-10\right)^{2}\times 2=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3k-10\right)^{2}.

$$-48+15k+\left(9k^{2}-60k+100\right)\times 2=0$$

Use the distributive property to multiply 9k^{2}-60k+100 by 2.

$$-48+15k+18k^{2}-120k+200=0$$

Combine 15k and -120k to get -105k.

$$-48-105k+18k^{2}+200=0$$

Add -48 and 200 to get 152.

$$152-105k+18k^{2}=0$$

Subtract 152 from both sides. Anything subtracted from zero gives its negation.

$$-105k+18k^{2}=-152$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$18k^{2}-105k=-152$$

Divide both sides by 18.

$$\frac{18k^{2}-105k}{18}=-\frac{152}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$k^{2}+\left(-\frac{105}{18}\right)k=-\frac{152}{18}$$

Reduce the fraction \frac{-105}{18} to lowest terms by extracting and canceling out 3.

$$k^{2}-\frac{35}{6}k=-\frac{152}{18}$$

Reduce the fraction \frac{-152}{18} to lowest terms by extracting and canceling out 2.

$$k^{2}-\frac{35}{6}k=-\frac{76}{9}$$

Divide -\frac{35}{6}, the coefficient of the x term, by 2 to get -\frac{35}{12}. Then add the square of -\frac{35}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-\frac{35}{6}k+\left(-\frac{35}{12}\right)^{2}=-\frac{76}{9}+\left(-\frac{35}{12}\right)^{2}$$

Square -\frac{35}{12} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-\frac{35}{6}k+\frac{1225}{144}=-\frac{76}{9}+\frac{1225}{144}$$

Add -\frac{76}{9} to \frac{1225}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}-\frac{35}{6}k+\frac{1225}{144}=\frac{1}{16}$$

Factor k^{2}-\frac{35}{6}k+\frac{1225}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-\frac{35}{12}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{35}{12}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$k-\frac{35}{12}=\frac{1}{4} k-\frac{35}{12}=-\frac{1}{4}$$

Add \frac{35}{12} to both sides of the equation.

$$k=\frac{19}{6} k=\frac{8}{3}$$