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y\left(5y+10\right)=0
Factor out y.
y=0 y=-2
To find equation solutions, solve y=0 and 5y+10=0.
5y^{2}+10y=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{-10±\sqrt{10^{2}}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 10 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-10±10}{2\times 5}
Take the square root of 10^{2}.
y=\frac{-10±10}{10}
Multiply 2 times 5.
y=\frac{0}{10}
Now solve the equation y=\frac{-10±10}{10} when ± is plus. Add -10 to 10.
y=0
Divide 0 by 10.
y=-\frac{20}{10}
Now solve the equation y=\frac{-10±10}{10} when ± is minus. Subtract 10 from -10.
y=-2
Divide -20 by 10.
y=0 y=-2
The equation is now solved.
5y^{2}+10y=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{5y^{2}+10y}{5}=\frac{0}{5}
Divide both sides by 5.
y^{2}+\frac{10}{5}y=\frac{0}{5}
Dividing by 5 undoes the multiplication by 5.
y^{2}+2y=\frac{0}{5}
Divide 10 by 5.
y^{2}+2y=0
Divide 0 by 5.
y^{2}+2y+1^{2}=1^{2}
Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}+2y+1=1
Square 1.
\left(y+1\right)^{2}=1
Factor y^{2}+2y+1. 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+1\right)^{2}}=\sqrt{1}
Take the square root of both sides of the equation.
y+1=1 y+1=-1
Simplify.
y=0 y=-2
Subtract 1 from both sides of the equation.