Solve (15-10y)(1+y)=12

$$( 15 - 10 y ) ( 1 + y ) = 12$$

Solve for y

$$y=\frac{\sqrt{145}}{20}+\frac{1}{4}\approx 0.852079729$$

$$y=-\frac{\sqrt{145}}{20}+\frac{1}{4}\approx -0.352079729$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 15-10y by 1+y and combine like terms.

$$15+5y-10y^{2}=12$$

Subtract 12 from both sides.

$$15+5y-10y^{2}-12=0$$

Subtract 12 from 15 to get 3.

$$3+5y-10y^{2}=0$$

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.

$$-10y^{2}+5y+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 5 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-5±\sqrt{5^{2}-4\left(-10\right)\times 3}}{2\left(-10\right)}$$

Square 5.

$$y=\frac{-5±\sqrt{25-4\left(-10\right)\times 3}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$y=\frac{-5±\sqrt{25+40\times 3}}{2\left(-10\right)}$$

Multiply 40 times 3.

$$y=\frac{-5±\sqrt{25+120}}{2\left(-10\right)}$$

Add 25 to 120.

$$y=\frac{-5±\sqrt{145}}{2\left(-10\right)}$$

Multiply 2 times -10.

$$y=\frac{-5±\sqrt{145}}{-20}$$

Now solve the equation y=\frac{-5±\sqrt{145}}{-20} when ± is plus. Add -5 to \sqrt{145}.

$$y=\frac{\sqrt{145}-5}{-20}$$

Divide -5+\sqrt{145} by -20.

$$y=-\frac{\sqrt{145}}{20}+\frac{1}{4}$$

Now solve the equation y=\frac{-5±\sqrt{145}}{-20} when ± is minus. Subtract \sqrt{145} from -5.

$$y=\frac{-\sqrt{145}-5}{-20}$$

Divide -5-\sqrt{145} by -20.

$$y=\frac{\sqrt{145}}{20}+\frac{1}{4}$$

The equation is now solved.

$$y=-\frac{\sqrt{145}}{20}+\frac{1}{4} y=\frac{\sqrt{145}}{20}+\frac{1}{4}$$

Steps for Completing the Square

Use the distributive property to multiply 15-10y by 1+y and combine like terms.

$$15+5y-10y^{2}=12$$

Subtract 15 from both sides.

$$5y-10y^{2}=12-15$$

Subtract 15 from 12 to get -3.

$$5y-10y^{2}=-3$$

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.

$$-10y^{2}+5y=-3$$

Divide both sides by -10.

$$\frac{-10y^{2}+5y}{-10}=-\frac{3}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$y^{2}+\frac{5}{-10}y=-\frac{3}{-10}$$

Reduce the fraction \frac{5}{-10} to lowest terms by extracting and canceling out 5.

$$y^{2}-\frac{1}{2}y=-\frac{3}{-10}$$

Divide -3 by -10.

$$y^{2}-\frac{1}{2}y=\frac{3}{10}$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{1}{2}y+\left(-\frac{1}{4}\right)^{2}=\frac{3}{10}+\left(-\frac{1}{4}\right)^{2}$$

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

$$y^{2}-\frac{1}{2}y+\frac{1}{16}=\frac{3}{10}+\frac{1}{16}$$

Add \frac{3}{10} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{1}{2}y+\frac{1}{16}=\frac{29}{80}$$

Factor y^{2}-\frac{1}{2}y+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{1}{4}\right)^{2}=\frac{29}{80}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{1}{4}\right)^{2}}=\sqrt{\frac{29}{80}}$$

Simplify.

$$y-\frac{1}{4}=\frac{\sqrt{145}}{20} y-\frac{1}{4}=-\frac{\sqrt{145}}{20}$$

Add \frac{1}{4} to both sides of the equation.

$$y=\frac{\sqrt{145}}{20}+\frac{1}{4} y=-\frac{\sqrt{145}}{20}+\frac{1}{4}$$