Solve 20y^2-20y-5=y

$$20 y ^ { 2 } - 20 y - 5 = y$$

Solve for y

$$y=-\frac{1}{5}=-0.2$$

$$y = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

Steps Using Factoring By Grouping

Subtract y from both sides.

$$20y^{2}-20y-5-y=0$$

Combine -20y and -y to get -21y.

$$20y^{2}-21y-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 20y^{2}+ay+by-5. To find a and b, set up a system to be solved.

$$a+b=-21 ab=20\left(-5\right)=-100$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -100.

$$1,-100 2,-50 4,-25 5,-20 10,-10$$

Calculate the sum for each pair.

$$1-100=-99 2-50=-48 4-25=-21 5-20=-15 10-10=0$$

The solution is the pair that gives sum -21.

$$a=-25 b=4$$

Rewrite 20y^{2}-21y-5 as \left(20y^{2}-25y\right)+\left(4y-5\right).

$$\left(20y^{2}-25y\right)+\left(4y-5\right)$$

Factor out 5y in 20y^{2}-25y.

$$5y\left(4y-5\right)+4y-5$$

Factor out common term 4y-5 by using distributive property.

$$\left(4y-5\right)\left(5y+1\right)$$

To find equation solutions, solve 4y-5=0 and 5y+1=0.

$$y=\frac{5}{4} y=-\frac{1}{5}$$

Steps Using the Quadratic Formula

Subtract y from both sides.

$$20y^{2}-20y-5-y=0$$

Combine -20y and -y to get -21y.

$$20y^{2}-21y-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, -21 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 20\left(-5\right)}}{2\times 20}$$

Square -21.

$$y=\frac{-\left(-21\right)±\sqrt{441-4\times 20\left(-5\right)}}{2\times 20}$$

Multiply -4 times 20.

$$y=\frac{-\left(-21\right)±\sqrt{441-80\left(-5\right)}}{2\times 20}$$

Multiply -80 times -5.

$$y=\frac{-\left(-21\right)±\sqrt{441+400}}{2\times 20}$$

Add 441 to 400.

$$y=\frac{-\left(-21\right)±\sqrt{841}}{2\times 20}$$

Take the square root of 841.

$$y=\frac{-\left(-21\right)±29}{2\times 20}$$

The opposite of -21 is 21.

$$y=\frac{21±29}{2\times 20}$$

Multiply 2 times 20.

$$y=\frac{21±29}{40}$$

Now solve the equation y=\frac{21±29}{40} when ± is plus. Add 21 to 29.

$$y=\frac{50}{40}$$

Reduce the fraction \frac{50}{40} to lowest terms by extracting and canceling out 10.

$$y=\frac{5}{4}$$

Now solve the equation y=\frac{21±29}{40} when ± is minus. Subtract 29 from 21.

$$y=-\frac{8}{40}$$

Reduce the fraction \frac{-8}{40} to lowest terms by extracting and canceling out 8.

$$y=-\frac{1}{5}$$

The equation is now solved.

$$y=\frac{5}{4} y=-\frac{1}{5}$$

Steps for Completing the Square

Subtract y from both sides.

$$20y^{2}-20y-5-y=0$$

Combine -20y and -y to get -21y.

$$20y^{2}-21y-5=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$20y^{2}-21y=5$$

Divide both sides by 20.

$$\frac{20y^{2}-21y}{20}=\frac{5}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$y^{2}-\frac{21}{20}y=\frac{5}{20}$$

Reduce the fraction \frac{5}{20} to lowest terms by extracting and canceling out 5.

$$y^{2}-\frac{21}{20}y=\frac{1}{4}$$

Divide -\frac{21}{20}, the coefficient of the x term, by 2 to get -\frac{21}{40}. Then add the square of -\frac{21}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{21}{20}y+\left(-\frac{21}{40}\right)^{2}=\frac{1}{4}+\left(-\frac{21}{40}\right)^{2}$$

Square -\frac{21}{40} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{21}{20}y+\frac{441}{1600}=\frac{1}{4}+\frac{441}{1600}$$

Add \frac{1}{4} to \frac{441}{1600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{21}{20}y+\frac{441}{1600}=\frac{841}{1600}$$

Factor y^{2}-\frac{21}{20}y+\frac{441}{1600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{21}{40}\right)^{2}=\frac{841}{1600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{21}{40}\right)^{2}}=\sqrt{\frac{841}{1600}}$$

Simplify.

$$y-\frac{21}{40}=\frac{29}{40} y-\frac{21}{40}=-\frac{29}{40}$$

Add \frac{21}{40} to both sides of the equation.

$$y=\frac{5}{4} y=-\frac{1}{5}$$