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y\left(4y+28\right)=0
Factor out y.
y=0 y=-7
To find equation solutions, solve y=0 and 4y+28=0.
4y^{2}+28y=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{-28±\sqrt{28^{2}}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 28 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-28±28}{2\times 4}
Take the square root of 28^{2}.
y=\frac{-28±28}{8}
Multiply 2 times 4.
y=\frac{0}{8}
Now solve the equation y=\frac{-28±28}{8} when ± is plus. Add -28 to 28.
y=0
Divide 0 by 8.
y=-\frac{56}{8}
Now solve the equation y=\frac{-28±28}{8} when ± is minus. Subtract 28 from -28.
y=-7
Divide -56 by 8.
y=0 y=-7
The equation is now solved.
4y^{2}+28y=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{4y^{2}+28y}{4}=\frac{0}{4}
Divide both sides by 4.
y^{2}+\frac{28}{4}y=\frac{0}{4}
Dividing by 4 undoes the multiplication by 4.
y^{2}+7y=\frac{0}{4}
Divide 28 by 4.
y^{2}+7y=0
Divide 0 by 4.
y^{2}+7y+\left(\frac{7}{2}\right)^{2}=\left(\frac{7}{2}\right)^{2}
Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}+7y+\frac{49}{4}=\frac{49}{4}
Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.
\left(y+\frac{7}{2}\right)^{2}=\frac{49}{4}
Factor y^{2}+7y+\frac{49}{4}. 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{7}{2}\right)^{2}}=\sqrt{\frac{49}{4}}
Take the square root of both sides of the equation.
y+\frac{7}{2}=\frac{7}{2} y+\frac{7}{2}=-\frac{7}{2}
Simplify.
y=0 y=-7
Subtract \frac{7}{2} from both sides of the equation.