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s\left(s+12\right)=0
Factor out s.
s=0 s=-12
To find equation solutions, solve s=0 and s+12=0.
s^{2}+12s=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.
s=\frac{-12±\sqrt{12^{2}}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 12 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
s=\frac{-12±12}{2}
Take the square root of 12^{2}.
s=\frac{0}{2}
Now solve the equation s=\frac{-12±12}{2} when ± is plus. Add -12 to 12.
s=0
Divide 0 by 2.
s=-\frac{24}{2}
Now solve the equation s=\frac{-12±12}{2} when ± is minus. Subtract 12 from -12.
s=-12
Divide -24 by 2.
s=0 s=-12
The equation is now solved.
s^{2}+12s=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.
s^{2}+12s+6^{2}=6^{2}
Divide 12, the coefficient of the x term, by 2 to get 6. Then add the square of 6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
s^{2}+12s+36=36
Square 6.
\left(s+6\right)^{2}=36
Factor s^{2}+12s+36. 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(s+6\right)^{2}}=\sqrt{36}
Take the square root of both sides of the equation.
s+6=6 s+6=-6
Simplify.
s=0 s=-12
Subtract 6 from both sides of the equation.