Solve 9y^2-30y=-25

$$9 y ^ { 2 } - 30 y = - 25$$

Solve for y

$$y = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

Steps Using Factoring By Grouping

Add 25 to both sides.

$$9y^{2}-30y+25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9y^{2}+ay+by+25. To find a and b, set up a system to be solved.

$$a+b=-30 ab=9\times 25=225$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 225.

$$-1,-225 -3,-75 -5,-45 -9,-25 -15,-15$$

Calculate the sum for each pair.

$$-1-225=-226 -3-75=-78 -5-45=-50 -9-25=-34 -15-15=-30$$

The solution is the pair that gives sum -30.

$$a=-15 b=-15$$

Rewrite 9y^{2}-30y+25 as \left(9y^{2}-15y\right)+\left(-15y+25\right).

$$\left(9y^{2}-15y\right)+\left(-15y+25\right)$$

Factor out 3y in the first and -5 in the second group.

$$3y\left(3y-5\right)-5\left(3y-5\right)$$

Factor out common term 3y-5 by using distributive property.

$$\left(3y-5\right)\left(3y-5\right)$$

Rewrite as a binomial square.

$$\left(3y-5\right)^{2}$$

To find equation solution, solve 3y-5=0.

$$y=\frac{5}{3}$$

Steps Using the Quadratic Formula

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.

$$9y^{2}-30y=-25$$

Add 25 to both sides of the equation.

$$9y^{2}-30y-\left(-25\right)=-25-\left(-25\right)$$

Subtracting -25 from itself leaves 0.

$$9y^{2}-30y-\left(-25\right)=0$$

Subtract -25 from 0.

$$9y^{2}-30y+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -30 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 9\times 25}}{2\times 9}$$

Square -30.

$$y=\frac{-\left(-30\right)±\sqrt{900-4\times 9\times 25}}{2\times 9}$$

Multiply -4 times 9.

$$y=\frac{-\left(-30\right)±\sqrt{900-36\times 25}}{2\times 9}$$

Multiply -36 times 25.

$$y=\frac{-\left(-30\right)±\sqrt{900-900}}{2\times 9}$$

Add 900 to -900.

$$y=\frac{-\left(-30\right)±\sqrt{0}}{2\times 9}$$

Take the square root of 0.

$$y=-\frac{-30}{2\times 9}$$

The opposite of -30 is 30.

$$y=\frac{30}{2\times 9}$$

Multiply 2 times 9.

$$y=\frac{30}{18}$$

Reduce the fraction \frac{30}{18} to lowest terms by extracting and canceling out 6.

$$y=\frac{5}{3}$$

Steps for Completing the Square

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.

$$9y^{2}-30y=-25$$

Divide both sides by 9.

$$\frac{9y^{2}-30y}{9}=-\frac{25}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$y^{2}+\left(-\frac{30}{9}\right)y=-\frac{25}{9}$$

Reduce the fraction \frac{-30}{9} to lowest terms by extracting and canceling out 3.

$$y^{2}-\frac{10}{3}y=-\frac{25}{9}$$

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{10}{3}y+\left(-\frac{5}{3}\right)^{2}=-\frac{25}{9}+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{10}{3}y+\frac{25}{9}=\frac{-25+25}{9}$$

Add -\frac{25}{9} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{10}{3}y+\frac{25}{9}=0$$

Factor y^{2}-\frac{10}{3}y+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{5}{3}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{5}{3}\right)^{2}}=\sqrt{0}$$

Simplify.

$$y-\frac{5}{3}=0 y-\frac{5}{3}=0$$

Add \frac{5}{3} to both sides of the equation.

$$y=\frac{5}{3} y=\frac{5}{3}$$

The equation is now solved. Solutions are the same.

$$y=\frac{5}{3}$$