Solve 17/frac{20y{y-20}}+8y=0.6

$$\frac{ 17 }{ \frac{ 20y }{ y-20 } } +8y=0.6$$

Solve for y

$$y = \frac{\sqrt{8705} - 1}{64} \approx 1.442196711$$

$$y=\frac{-\sqrt{8705}-1}{64}\approx -1.473446711$$

Steps Using the Quadratic Formula

Variable y cannot be equal to 20 since division by zero is not defined. Divide 17 by \frac{20y}{y-20} by multiplying 17 by the reciprocal of \frac{20y}{y-20}.

$$\frac{17\left(y-20\right)}{20y}+8y=0.6$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 8y times \frac{20y}{20y}.

$$\frac{17\left(y-20\right)}{20y}+\frac{8y\times 20y}{20y}=0.6$$

Since \frac{17\left(y-20\right)}{20y} and \frac{8y\times 20y}{20y} have the same denominator, add them by adding their numerators.

$$\frac{17\left(y-20\right)+8y\times 20y}{20y}=0.6$$

Do the multiplications in 17\left(y-20\right)+8y\times 20y.

$$\frac{17y-340+160y^{2}}{20y}=0.6$$

Factor the expressions that are not already factored in \frac{17y-340+160y^{2}}{20y}.

$$\frac{160\left(y-\left(-\frac{1}{320}\sqrt{217889}-\frac{17}{320}\right)\right)\left(y-\left(\frac{1}{320}\sqrt{217889}-\frac{17}{320}\right)\right)}{20y}=0.6$$

Cancel out 20 in both numerator and denominator.

$$\frac{8\left(y-\left(-\frac{1}{320}\sqrt{217889}-\frac{17}{320}\right)\right)\left(y-\left(\frac{1}{320}\sqrt{217889}-\frac{17}{320}\right)\right)}{y}=0.6$$

To find the opposite of -\frac{1}{320}\sqrt{217889}-\frac{17}{320}, find the opposite of each term.

$$\frac{8\left(y+\frac{1}{320}\sqrt{217889}+\frac{17}{320}\right)\left(y-\left(\frac{1}{320}\sqrt{217889}-\frac{17}{320}\right)\right)}{y}=0.6$$

To find the opposite of \frac{1}{320}\sqrt{217889}-\frac{17}{320}, find the opposite of each term.

$$\frac{8\left(y+\frac{1}{320}\sqrt{217889}+\frac{17}{320}\right)\left(y-\frac{1}{320}\sqrt{217889}+\frac{17}{320}\right)}{y}=0.6$$

Use the distributive property to multiply 8 by y+\frac{1}{320}\sqrt{217889}+\frac{17}{320}.

$$\frac{\left(8y+\frac{1}{40}\sqrt{217889}+\frac{17}{40}\right)\left(y-\frac{1}{320}\sqrt{217889}+\frac{17}{320}\right)}{y}=0.6$$

Use the distributive property to multiply 8y+\frac{1}{40}\sqrt{217889}+\frac{17}{40} by y-\frac{1}{320}\sqrt{217889}+\frac{17}{320} and combine like terms.

$$\frac{8y^{2}+\frac{17}{20}y-\frac{1}{12800}\left(\sqrt{217889}\right)^{2}+\frac{289}{12800}}{y}=0.6$$

The square of \sqrt{217889} is 217889.

$$\frac{8y^{2}+\frac{17}{20}y-\frac{1}{12800}\times 217889+\frac{289}{12800}}{y}=0.6$$

Multiply -\frac{1}{12800} and 217889 to get -\frac{217889}{12800}.

$$\frac{8y^{2}+\frac{17}{20}y-\frac{217889}{12800}+\frac{289}{12800}}{y}=0.6$$

Add -\frac{217889}{12800} and \frac{289}{12800} to get -17.

$$\frac{8y^{2}+\frac{17}{20}y-17}{y}=0.6$$

Subtract 0.6 from both sides.

$$\frac{8y^{2}+\frac{17}{20}y-17}{y}-0.6=0$$

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y.

$$8y^{2}+\frac{17}{20}y-17+y\left(-0.6\right)=0$$

Reorder the terms.

$$8y^{2}-0.6y+\frac{17}{20}y-17=0$$

Combine -0.6y and \frac{17}{20}y to get \frac{1}{4}y.

$$8y^{2}+\frac{1}{4}y-17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, \frac{1}{4} for b, and -17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\frac{1}{4}±\sqrt{\left(\frac{1}{4}\right)^{2}-4\times 8\left(-17\right)}}{2\times 8}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$y=\frac{-\frac{1}{4}±\sqrt{\frac{1}{16}-4\times 8\left(-17\right)}}{2\times 8}$$

Multiply -4 times 8.

$$y=\frac{-\frac{1}{4}±\sqrt{\frac{1}{16}-32\left(-17\right)}}{2\times 8}$$

Multiply -32 times -17.

$$y=\frac{-\frac{1}{4}±\sqrt{\frac{1}{16}+544}}{2\times 8}$$

Add \frac{1}{16} to 544.

$$y=\frac{-\frac{1}{4}±\sqrt{\frac{8705}{16}}}{2\times 8}$$

Take the square root of \frac{8705}{16}.

$$y=\frac{-\frac{1}{4}±\frac{\sqrt{8705}}{4}}{2\times 8}$$

Multiply 2 times 8.

$$y=\frac{-\frac{1}{4}±\frac{\sqrt{8705}}{4}}{16}$$

Now solve the equation y=\frac{-\frac{1}{4}±\frac{\sqrt{8705}}{4}}{16} when ± is plus. Add -\frac{1}{4} to \frac{\sqrt{8705}}{4}.

$$y=\frac{\sqrt{8705}-1}{4\times 16}$$

Divide \frac{-1+\sqrt{8705}}{4} by 16.

$$y=\frac{\sqrt{8705}-1}{64}$$

Now solve the equation y=\frac{-\frac{1}{4}±\frac{\sqrt{8705}}{4}}{16} when ± is minus. Subtract \frac{\sqrt{8705}}{4} from -\frac{1}{4}.

$$y=\frac{-\sqrt{8705}-1}{4\times 16}$$

Divide \frac{-1-\sqrt{8705}}{4} by 16.

$$y=\frac{-\sqrt{8705}-1}{64}$$

The equation is now solved.

$$y=\frac{\sqrt{8705}-1}{64} y=\frac{-\sqrt{8705}-1}{64}$$

Steps for Completing the Square

Variable y cannot be equal to 20 since division by zero is not defined. Divide 17 by \frac{20y}{y-20} by multiplying 17 by the reciprocal of \frac{20y}{y-20}.

$$\frac{17\left(y-20\right)}{20y}+8y=0.6$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 8y times \frac{20y}{20y}.

$$\frac{17\left(y-20\right)}{20y}+\frac{8y\times 20y}{20y}=0.6$$

Since \frac{17\left(y-20\right)}{20y} and \frac{8y\times 20y}{20y} have the same denominator, add them by adding their numerators.

$$\frac{17\left(y-20\right)+8y\times 20y}{20y}=0.6$$

Do the multiplications in 17\left(y-20\right)+8y\times 20y.

$$\frac{17y-340+160y^{2}}{20y}=0.6$$

Factor the expressions that are not already factored in \frac{17y-340+160y^{2}}{20y}.

$$\frac{160\left(y-\left(-\frac{1}{320}\sqrt{217889}-\frac{17}{320}\right)\right)\left(y-\left(\frac{1}{320}\sqrt{217889}-\frac{17}{320}\right)\right)}{20y}=0.6$$

Cancel out 20 in both numerator and denominator.

$$\frac{8\left(y-\left(-\frac{1}{320}\sqrt{217889}-\frac{17}{320}\right)\right)\left(y-\left(\frac{1}{320}\sqrt{217889}-\frac{17}{320}\right)\right)}{y}=0.6$$

To find the opposite of -\frac{1}{320}\sqrt{217889}-\frac{17}{320}, find the opposite of each term.

$$\frac{8\left(y+\frac{1}{320}\sqrt{217889}+\frac{17}{320}\right)\left(y-\left(\frac{1}{320}\sqrt{217889}-\frac{17}{320}\right)\right)}{y}=0.6$$

To find the opposite of \frac{1}{320}\sqrt{217889}-\frac{17}{320}, find the opposite of each term.

$$\frac{8\left(y+\frac{1}{320}\sqrt{217889}+\frac{17}{320}\right)\left(y-\frac{1}{320}\sqrt{217889}+\frac{17}{320}\right)}{y}=0.6$$

Use the distributive property to multiply 8 by y+\frac{1}{320}\sqrt{217889}+\frac{17}{320}.

$$\frac{\left(8y+\frac{1}{40}\sqrt{217889}+\frac{17}{40}\right)\left(y-\frac{1}{320}\sqrt{217889}+\frac{17}{320}\right)}{y}=0.6$$

Use the distributive property to multiply 8y+\frac{1}{40}\sqrt{217889}+\frac{17}{40} by y-\frac{1}{320}\sqrt{217889}+\frac{17}{320} and combine like terms.

$$\frac{8y^{2}+\frac{17}{20}y-\frac{1}{12800}\left(\sqrt{217889}\right)^{2}+\frac{289}{12800}}{y}=0.6$$

The square of \sqrt{217889} is 217889.

$$\frac{8y^{2}+\frac{17}{20}y-\frac{1}{12800}\times 217889+\frac{289}{12800}}{y}=0.6$$

Multiply -\frac{1}{12800} and 217889 to get -\frac{217889}{12800}.

$$\frac{8y^{2}+\frac{17}{20}y-\frac{217889}{12800}+\frac{289}{12800}}{y}=0.6$$

Add -\frac{217889}{12800} and \frac{289}{12800} to get -17.

$$\frac{8y^{2}+\frac{17}{20}y-17}{y}=0.6$$

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y.

$$8y^{2}+\frac{17}{20}y-17=0.6y$$

Subtract 0.6y from both sides.

$$8y^{2}+\frac{17}{20}y-17-0.6y=0$$

Combine \frac{17}{20}y and -0.6y to get \frac{1}{4}y.

$$8y^{2}+\frac{1}{4}y-17=0$$

Add 17 to both sides. Anything plus zero gives itself.

$$8y^{2}+\frac{1}{4}y=17$$

Divide both sides by 8.

$$\frac{8y^{2}+\frac{1}{4}y}{8}=\frac{17}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$y^{2}+\frac{\frac{1}{4}}{8}y=\frac{17}{8}$$

Divide \frac{1}{4} by 8.

$$y^{2}+\frac{1}{32}y=\frac{17}{8}$$

Divide \frac{1}{32}, the coefficient of the x term, by 2 to get \frac{1}{64}. Then add the square of \frac{1}{64} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+\frac{1}{32}y+\left(\frac{1}{64}\right)^{2}=\frac{17}{8}+\left(\frac{1}{64}\right)^{2}$$

Square \frac{1}{64} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+\frac{1}{32}y+\frac{1}{4096}=\frac{17}{8}+\frac{1}{4096}$$

Add \frac{17}{8} to \frac{1}{4096} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}+\frac{1}{32}y+\frac{1}{4096}=\frac{8705}{4096}$$

Factor y^{2}+\frac{1}{32}y+\frac{1}{4096}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+\frac{1}{64}\right)^{2}=\frac{8705}{4096}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{1}{64}\right)^{2}}=\sqrt{\frac{8705}{4096}}$$

Simplify.

$$y+\frac{1}{64}=\frac{\sqrt{8705}}{64} y+\frac{1}{64}=-\frac{\sqrt{8705}}{64}$$

Subtract \frac{1}{64} from both sides of the equation.

$$y=\frac{\sqrt{8705}-1}{64} y=\frac{-\sqrt{8705}-1}{64}$$