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12y-8y^{2}=0
Subtract 8y^{2} from both sides.
y\left(12-8y\right)=0
Factor out y.
y=0 y=\frac{3}{2}
To find equation solutions, solve y=0 and 12-8y=0.
12y-8y^{2}=0
Subtract 8y^{2} from both sides.
-8y^{2}+12y=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=\frac{-12±\sqrt{12^{2}}}{2\left(-8\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 12 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-12±12}{2\left(-8\right)}
Take the square root of 12^{2}.
y=\frac{-12±12}{-16}
Multiply 2 times -8.
y=\frac{0}{-16}
Now solve the equation y=\frac{-12±12}{-16} when ± is plus. Add -12 to 12.
y=0
Divide 0 by -16.
y=-\frac{24}{-16}
Now solve the equation y=\frac{-12±12}{-16} when ± is minus. Subtract 12 from -12.
y=\frac{3}{2}
Reduce the fraction \frac{-24}{-16} to lowest terms by extracting and canceling out 8.
y=0 y=\frac{3}{2}
The equation is now solved.
12y-8y^{2}=0
Subtract 8y^{2} from both sides.
-8y^{2}+12y=0
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.
\frac{-8y^{2}+12y}{-8}=\frac{0}{-8}
Divide both sides by -8.
y^{2}+\frac{12}{-8}y=\frac{0}{-8}
Dividing by -8 undoes the multiplication by -8.
y^{2}-\frac{3}{2}y=\frac{0}{-8}
Reduce the fraction \frac{12}{-8} to lowest terms by extracting and canceling out 4.
y^{2}-\frac{3}{2}y=0
Divide 0 by -8.
y^{2}-\frac{3}{2}y+\left(-\frac{3}{4}\right)^{2}=\left(-\frac{3}{4}\right)^{2}
Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}-\frac{3}{2}y+\frac{9}{16}=\frac{9}{16}
Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.
\left(y-\frac{3}{4}\right)^{2}=\frac{9}{16}
Factor y^{2}-\frac{3}{2}y+\frac{9}{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}.
\sqrt{\left(y-\frac{3}{4}\right)^{2}}=\sqrt{\frac{9}{16}}
Take the square root of both sides of the equation.
y-\frac{3}{4}=\frac{3}{4} y-\frac{3}{4}=-\frac{3}{4}
Simplify.
y=\frac{3}{2} y=0
Add \frac{3}{4} to both sides of the equation.