Solve 81/t=2/2-t+50

$$\frac { 81 } { t } = \frac { 2 } { 2 - t } + 50$$

Solve for t

$$t = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$t = \frac{54}{25} = 2\frac{4}{25} = 2.16$$

Steps Using the Quadratic Formula

Variable t cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by t\left(t-2\right), the least common multiple of t,2-t.

$$\left(t-2\right)\times 81=-t\times 2+t\left(t-2\right)\times 50$$

Use the distributive property to multiply t-2 by 81.

$$81t-162=-t\times 2+t\left(t-2\right)\times 50$$

Multiply -1 and 2 to get -2.

$$81t-162=-2t+t\left(t-2\right)\times 50$$

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

$$81t-162=-2t+\left(t^{2}-2t\right)\times 50$$

Use the distributive property to multiply t^{2}-2t by 50.

$$81t-162=-2t+50t^{2}-100t$$

Combine -2t and -100t to get -102t.

$$81t-162=-102t+50t^{2}$$

Add 102t to both sides.

$$81t-162+102t=50t^{2}$$

Combine 81t and 102t to get 183t.

$$183t-162=50t^{2}$$

Subtract 50t^{2} from both sides.

$$183t-162-50t^{2}=0$$

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.

$$-50t^{2}+183t-162=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -50 for a, 183 for b, and -162 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-183±\sqrt{183^{2}-4\left(-50\right)\left(-162\right)}}{2\left(-50\right)}$$

Square 183.

$$t=\frac{-183±\sqrt{33489-4\left(-50\right)\left(-162\right)}}{2\left(-50\right)}$$

Multiply -4 times -50.

$$t=\frac{-183±\sqrt{33489+200\left(-162\right)}}{2\left(-50\right)}$$

Multiply 200 times -162.

$$t=\frac{-183±\sqrt{33489-32400}}{2\left(-50\right)}$$

Add 33489 to -32400.

$$t=\frac{-183±\sqrt{1089}}{2\left(-50\right)}$$

Take the square root of 1089.

$$t=\frac{-183±33}{2\left(-50\right)}$$

Multiply 2 times -50.

$$t=\frac{-183±33}{-100}$$

Now solve the equation t=\frac{-183±33}{-100} when ± is plus. Add -183 to 33.

$$t=-\frac{150}{-100}$$

Reduce the fraction \frac{-150}{-100} to lowest terms by extracting and canceling out 50.

$$t=\frac{3}{2}$$

Now solve the equation t=\frac{-183±33}{-100} when ± is minus. Subtract 33 from -183.

$$t=-\frac{216}{-100}$$

Reduce the fraction \frac{-216}{-100} to lowest terms by extracting and canceling out 4.

$$t=\frac{54}{25}$$

The equation is now solved.

$$t=\frac{3}{2} t=\frac{54}{25}$$

Steps for Completing the Square

Variable t cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by t\left(t-2\right), the least common multiple of t,2-t.

$$\left(t-2\right)\times 81=-t\times 2+t\left(t-2\right)\times 50$$

Use the distributive property to multiply t-2 by 81.

$$81t-162=-t\times 2+t\left(t-2\right)\times 50$$

Multiply -1 and 2 to get -2.

$$81t-162=-2t+t\left(t-2\right)\times 50$$

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

$$81t-162=-2t+\left(t^{2}-2t\right)\times 50$$

Use the distributive property to multiply t^{2}-2t by 50.

$$81t-162=-2t+50t^{2}-100t$$

Combine -2t and -100t to get -102t.

$$81t-162=-102t+50t^{2}$$

Add 102t to both sides.

$$81t-162+102t=50t^{2}$$

Combine 81t and 102t to get 183t.

$$183t-162=50t^{2}$$

Subtract 50t^{2} from both sides.

$$183t-162-50t^{2}=0$$

Add 162 to both sides. Anything plus zero gives itself.

$$183t-50t^{2}=162$$

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.

$$-50t^{2}+183t=162$$

Divide both sides by -50.

$$\frac{-50t^{2}+183t}{-50}=\frac{162}{-50}$$

Dividing by -50 undoes the multiplication by -50.

$$t^{2}+\frac{183}{-50}t=\frac{162}{-50}$$

Divide 183 by -50.

$$t^{2}-\frac{183}{50}t=\frac{162}{-50}$$

Reduce the fraction \frac{162}{-50} to lowest terms by extracting and canceling out 2.

$$t^{2}-\frac{183}{50}t=-\frac{81}{25}$$

Divide -\frac{183}{50}, the coefficient of the x term, by 2 to get -\frac{183}{100}. Then add the square of -\frac{183}{100} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{183}{50}t+\left(-\frac{183}{100}\right)^{2}=-\frac{81}{25}+\left(-\frac{183}{100}\right)^{2}$$

Square -\frac{183}{100} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{183}{50}t+\frac{33489}{10000}=-\frac{81}{25}+\frac{33489}{10000}$$

Add -\frac{81}{25} to \frac{33489}{10000} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{183}{50}t+\frac{33489}{10000}=\frac{1089}{10000}$$

Factor t^{2}-\frac{183}{50}t+\frac{33489}{10000}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{183}{100}\right)^{2}=\frac{1089}{10000}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{183}{100}\right)^{2}}=\sqrt{\frac{1089}{10000}}$$

Simplify.

$$t-\frac{183}{100}=\frac{33}{100} t-\frac{183}{100}=-\frac{33}{100}$$

Add \frac{183}{100} to both sides of the equation.

$$t=\frac{54}{25} t=\frac{3}{2}$$