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x\left(5x+20\right)=0
Factor out x.
x=0 x=-4
To find equation solutions, solve x=0 and 5x+20=0.
5x^{2}+20x=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.
x=\frac{-20±\sqrt{20^{2}}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 20 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-20±20}{2\times 5}
Take the square root of 20^{2}.
x=\frac{-20±20}{10}
Multiply 2 times 5.
x=\frac{0}{10}
Now solve the equation x=\frac{-20±20}{10} when ± is plus. Add -20 to 20.
x=0
Divide 0 by 10.
x=-\frac{40}{10}
Now solve the equation x=\frac{-20±20}{10} when ± is minus. Subtract 20 from -20.
x=-4
Divide -40 by 10.
x=0 x=-4
The equation is now solved.
5x^{2}+20x=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{5x^{2}+20x}{5}=\frac{0}{5}
Divide both sides by 5.
x^{2}+\frac{20}{5}x=\frac{0}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}+4x=\frac{0}{5}
Divide 20 by 5.
x^{2}+4x=0
Divide 0 by 5.
x^{2}+4x+2^{2}=2^{2}
Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+4x+4=4
Square 2.
\left(x+2\right)^{2}=4
Factor x^{2}+4x+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(x+2\right)^{2}}=\sqrt{4}
Take the square root of both sides of the equation.
x+2=2 x+2=-2
Simplify.
x=0 x=-4
Subtract 2 from both sides of the equation.