Solve 2y^2+1/5-y=3(1/5-y)^2-2

$$2 y ^ { 2 } + \frac { 1 } { 5 } - y = 3 ( \frac { 1 } { 5 } - y ) ^ { 2 } - 2$$

Solve for y

$$y = \frac{\sqrt{209} + 1}{10} \approx 1.545683229$$

$$y=\frac{1-\sqrt{209}}{10}\approx -1.345683229$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{5}-y\right)^{2}.

$$2y^{2}+\frac{1}{5}-y=3\left(\frac{1}{25}-\frac{2}{5}y+y^{2}\right)-2$$

Use the distributive property to multiply 3 by \frac{1}{25}-\frac{2}{5}y+y^{2}.

$$2y^{2}+\frac{1}{5}-y=\frac{3}{25}-\frac{6}{5}y+3y^{2}-2$$

Subtract 2 from \frac{3}{25} to get -\frac{47}{25}.

$$2y^{2}+\frac{1}{5}-y=-\frac{47}{25}-\frac{6}{5}y+3y^{2}$$

Subtract -\frac{47}{25} from both sides.

$$2y^{2}+\frac{1}{5}-y-\left(-\frac{47}{25}\right)=-\frac{6}{5}y+3y^{2}$$

The opposite of -\frac{47}{25} is \frac{47}{25}.

$$2y^{2}+\frac{1}{5}-y+\frac{47}{25}=-\frac{6}{5}y+3y^{2}$$

Add \frac{6}{5}y to both sides.

$$2y^{2}+\frac{1}{5}-y+\frac{47}{25}+\frac{6}{5}y=3y^{2}$$

Add \frac{1}{5} and \frac{47}{25} to get \frac{52}{25}.

$$2y^{2}+\frac{52}{25}-y+\frac{6}{5}y=3y^{2}$$

Combine -y and \frac{6}{5}y to get \frac{1}{5}y.

$$2y^{2}+\frac{52}{25}+\frac{1}{5}y=3y^{2}$$

Subtract 3y^{2} from both sides.

$$2y^{2}+\frac{52}{25}+\frac{1}{5}y-3y^{2}=0$$

Combine 2y^{2} and -3y^{2} to get -y^{2}.

$$-y^{2}+\frac{52}{25}+\frac{1}{5}y=0$$

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.

$$-y^{2}+\frac{1}{5}y+\frac{52}{25}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, \frac{1}{5} for b, and \frac{52}{25} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\frac{1}{5}±\sqrt{\left(\frac{1}{5}\right)^{2}-4\left(-1\right)\times \frac{52}{25}}}{2\left(-1\right)}$$

Square \frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$y=\frac{-\frac{1}{5}±\sqrt{\frac{1}{25}-4\left(-1\right)\times \frac{52}{25}}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$y=\frac{-\frac{1}{5}±\sqrt{\frac{1}{25}+4\times \frac{52}{25}}}{2\left(-1\right)}$$

Multiply 4 times \frac{52}{25}.

$$y=\frac{-\frac{1}{5}±\sqrt{\frac{1+208}{25}}}{2\left(-1\right)}$$

Add \frac{1}{25} to \frac{208}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\frac{1}{5}±\sqrt{\frac{209}{25}}}{2\left(-1\right)}$$

Take the square root of \frac{209}{25}.

$$y=\frac{-\frac{1}{5}±\frac{\sqrt{209}}{5}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$y=\frac{-\frac{1}{5}±\frac{\sqrt{209}}{5}}{-2}$$

Now solve the equation y=\frac{-\frac{1}{5}±\frac{\sqrt{209}}{5}}{-2} when ± is plus. Add -\frac{1}{5} to \frac{\sqrt{209}}{5}.

$$y=\frac{\sqrt{209}-1}{-2\times 5}$$

Divide \frac{-1+\sqrt{209}}{5} by -2.

$$y=\frac{1-\sqrt{209}}{10}$$

Now solve the equation y=\frac{-\frac{1}{5}±\frac{\sqrt{209}}{5}}{-2} when ± is minus. Subtract \frac{\sqrt{209}}{5} from -\frac{1}{5}.

$$y=\frac{-\sqrt{209}-1}{-2\times 5}$$

Divide \frac{-1-\sqrt{209}}{5} by -2.

$$y=\frac{\sqrt{209}+1}{10}$$

The equation is now solved.

$$y=\frac{1-\sqrt{209}}{10} y=\frac{\sqrt{209}+1}{10}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{5}-y\right)^{2}.

$$2y^{2}+\frac{1}{5}-y=3\left(\frac{1}{25}-\frac{2}{5}y+y^{2}\right)-2$$

Use the distributive property to multiply 3 by \frac{1}{25}-\frac{2}{5}y+y^{2}.

$$2y^{2}+\frac{1}{5}-y=\frac{3}{25}-\frac{6}{5}y+3y^{2}-2$$

Subtract 2 from \frac{3}{25} to get -\frac{47}{25}.

$$2y^{2}+\frac{1}{5}-y=-\frac{47}{25}-\frac{6}{5}y+3y^{2}$$

Add \frac{6}{5}y to both sides.

$$2y^{2}+\frac{1}{5}-y+\frac{6}{5}y=-\frac{47}{25}+3y^{2}$$

Combine -y and \frac{6}{5}y to get \frac{1}{5}y.

$$2y^{2}+\frac{1}{5}+\frac{1}{5}y=-\frac{47}{25}+3y^{2}$$

Subtract 3y^{2} from both sides.

$$2y^{2}+\frac{1}{5}+\frac{1}{5}y-3y^{2}=-\frac{47}{25}$$

Combine 2y^{2} and -3y^{2} to get -y^{2}.

$$-y^{2}+\frac{1}{5}+\frac{1}{5}y=-\frac{47}{25}$$

Subtract \frac{1}{5} from both sides.

$$-y^{2}+\frac{1}{5}y=-\frac{47}{25}-\frac{1}{5}$$

Subtract \frac{1}{5} from -\frac{47}{25} to get -\frac{52}{25}.

$$-y^{2}+\frac{1}{5}y=-\frac{52}{25}$$

Divide both sides by -1.

$$\frac{-y^{2}+\frac{1}{5}y}{-1}=-\frac{\frac{52}{25}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$y^{2}+\frac{\frac{1}{5}}{-1}y=-\frac{\frac{52}{25}}{-1}$$

Divide \frac{1}{5} by -1.

$$y^{2}-\frac{1}{5}y=-\frac{\frac{52}{25}}{-1}$$

Divide -\frac{52}{25} by -1.

$$y^{2}-\frac{1}{5}y=\frac{52}{25}$$

Divide -\frac{1}{5}, the coefficient of the x term, by 2 to get -\frac{1}{10}. Then add the square of -\frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{1}{5}y+\left(-\frac{1}{10}\right)^{2}=\frac{52}{25}+\left(-\frac{1}{10}\right)^{2}$$

Square -\frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{1}{5}y+\frac{1}{100}=\frac{52}{25}+\frac{1}{100}$$

Add \frac{52}{25} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{1}{5}y+\frac{1}{100}=\frac{209}{100}$$

Factor y^{2}-\frac{1}{5}y+\frac{1}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{1}{10}\right)^{2}=\frac{209}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{1}{10}\right)^{2}}=\sqrt{\frac{209}{100}}$$

Simplify.

$$y-\frac{1}{10}=\frac{\sqrt{209}}{10} y-\frac{1}{10}=-\frac{\sqrt{209}}{10}$$

Add \frac{1}{10} to both sides of the equation.

$$y=\frac{\sqrt{209}+1}{10} y=\frac{1-\sqrt{209}}{10}$$