Solve -(9y-5)(3y+4)=0

$$- ( 9 y - 5 ) ( 3 y + 4 ) = 0$$

Solve for y

$$y = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

$$y=\frac{5}{9}\approx 0.555555556$$

Steps Using Factoring By Grouping

To find the opposite of 9y-5, find the opposite of each term.

$$\left(-9y-\left(-5\right)\right)\left(3y+4\right)=0$$

The opposite of -5 is 5.

$$\left(-9y+5\right)\left(3y+4\right)=0$$

Apply the distributive property by multiplying each term of -9y+5 by each term of 3y+4.

$$-27y^{2}-36y+15y+20=0$$

Combine -36y and 15y to get -21y.

$$-27y^{2}-21y+20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -27y^{2}+ay+by+20. To find a and b, set up a system to be solved.

$$a+b=-21 ab=-27\times 20=-540$$

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 -540.

$$1,-540 2,-270 3,-180 4,-135 5,-108 6,-90 9,-60 10,-54 12,-45 15,-36 18,-30 20,-27$$

Calculate the sum for each pair.

$$1-540=-539 2-270=-268 3-180=-177 4-135=-131 5-108=-103 6-90=-84 9-60=-51 10-54=-44 12-45=-33 15-36=-21 18-30=-12 20-27=-7$$

The solution is the pair that gives sum -21.

$$a=15 b=-36$$

Rewrite -27y^{2}-21y+20 as \left(-27y^{2}+15y\right)+\left(-36y+20\right).

$$\left(-27y^{2}+15y\right)+\left(-36y+20\right)$$

Factor out 3y in the first and 4 in the second group.

$$3y\left(-9y+5\right)+4\left(-9y+5\right)$$

Factor out common term -9y+5 by using distributive property.

$$\left(-9y+5\right)\left(3y+4\right)$$

To find equation solutions, solve -9y+5=0 and 3y+4=0.

$$y=\frac{5}{9} y=-\frac{4}{3}$$

Steps Using the Quadratic Formula

To find the opposite of 9y-5, find the opposite of each term.

$$\left(-9y-\left(-5\right)\right)\left(3y+4\right)=0$$

The opposite of -5 is 5.

$$\left(-9y+5\right)\left(3y+4\right)=0$$

Apply the distributive property by multiplying each term of -9y+5 by each term of 3y+4.

$$-27y^{2}-36y+15y+20=0$$

Combine -36y and 15y to get -21y.

$$-27y^{2}-21y+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -27 for a, -21 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\left(-27\right)\times 20}}{2\left(-27\right)}$$

Square -21.

$$y=\frac{-\left(-21\right)±\sqrt{441-4\left(-27\right)\times 20}}{2\left(-27\right)}$$

Multiply -4 times -27.

$$y=\frac{-\left(-21\right)±\sqrt{441+108\times 20}}{2\left(-27\right)}$$

Multiply 108 times 20.

$$y=\frac{-\left(-21\right)±\sqrt{441+2160}}{2\left(-27\right)}$$

Add 441 to 2160.

$$y=\frac{-\left(-21\right)±\sqrt{2601}}{2\left(-27\right)}$$

Take the square root of 2601.

$$y=\frac{-\left(-21\right)±51}{2\left(-27\right)}$$

The opposite of -21 is 21.

$$y=\frac{21±51}{2\left(-27\right)}$$

Multiply 2 times -27.

$$y=\frac{21±51}{-54}$$

Now solve the equation y=\frac{21±51}{-54} when ± is plus. Add 21 to 51.

$$y=\frac{72}{-54}$$

Reduce the fraction \frac{72}{-54} to lowest terms by extracting and canceling out 18.

$$y=-\frac{4}{3}$$

Now solve the equation y=\frac{21±51}{-54} when ± is minus. Subtract 51 from 21.

$$y=-\frac{30}{-54}$$

Reduce the fraction \frac{-30}{-54} to lowest terms by extracting and canceling out 6.

$$y=\frac{5}{9}$$

The equation is now solved.

$$y=-\frac{4}{3} y=\frac{5}{9}$$

Steps for Completing the Square

To find the opposite of 9y-5, find the opposite of each term.

$$\left(-9y-\left(-5\right)\right)\left(3y+4\right)=0$$

The opposite of -5 is 5.

$$\left(-9y+5\right)\left(3y+4\right)=0$$

Apply the distributive property by multiplying each term of -9y+5 by each term of 3y+4.

$$-27y^{2}-36y+15y+20=0$$

Combine -36y and 15y to get -21y.

$$-27y^{2}-21y+20=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$-27y^{2}-21y=-20$$

Divide both sides by -27.

$$\frac{-27y^{2}-21y}{-27}=-\frac{20}{-27}$$

Dividing by -27 undoes the multiplication by -27.

$$y^{2}+\left(-\frac{21}{-27}\right)y=-\frac{20}{-27}$$

Reduce the fraction \frac{-21}{-27} to lowest terms by extracting and canceling out 3.

$$y^{2}+\frac{7}{9}y=-\frac{20}{-27}$$

Divide -20 by -27.

$$y^{2}+\frac{7}{9}y=\frac{20}{27}$$

Divide \frac{7}{9}, the coefficient of the x term, by 2 to get \frac{7}{18}. Then add the square of \frac{7}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+\frac{7}{9}y+\left(\frac{7}{18}\right)^{2}=\frac{20}{27}+\left(\frac{7}{18}\right)^{2}$$

Square \frac{7}{18} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+\frac{7}{9}y+\frac{49}{324}=\frac{20}{27}+\frac{49}{324}$$

Add \frac{20}{27} to \frac{49}{324} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}+\frac{7}{9}y+\frac{49}{324}=\frac{289}{324}$$

Factor y^{2}+\frac{7}{9}y+\frac{49}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+\frac{7}{18}\right)^{2}=\frac{289}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{7}{18}\right)^{2}}=\sqrt{\frac{289}{324}}$$

Simplify.

$$y+\frac{7}{18}=\frac{17}{18} y+\frac{7}{18}=-\frac{17}{18}$$

Subtract \frac{7}{18} from both sides of the equation.

$$y=\frac{5}{9} y=-\frac{4}{3}$$