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s\left(s-5\right)=0
Factor out s.
s=0 s=5
To find equation solutions, solve s=0 and s-5=0.
s^{2}-5s=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{-\left(-5\right)±\sqrt{\left(-5\right)^{2}}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
s=\frac{-\left(-5\right)±5}{2}
Take the square root of \left(-5\right)^{2}.
s=\frac{5±5}{2}
The opposite of -5 is 5.
s=\frac{10}{2}
Now solve the equation s=\frac{5±5}{2} when ± is plus. Add 5 to 5.
s=5
Divide 10 by 2.
s=\frac{0}{2}
Now solve the equation s=\frac{5±5}{2} when ± is minus. Subtract 5 from 5.
s=0
Divide 0 by 2.
s=5 s=0
The equation is now solved.
s^{2}-5s=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}-5s+\left(-\frac{5}{2}\right)^{2}=\left(-\frac{5}{2}\right)^{2}
Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
s^{2}-5s+\frac{25}{4}=\frac{25}{4}
Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.
\left(s-\frac{5}{2}\right)^{2}=\frac{25}{4}
Factor s^{2}-5s+\frac{25}{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(s-\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}
Take the square root of both sides of the equation.
s-\frac{5}{2}=\frac{5}{2} s-\frac{5}{2}=-\frac{5}{2}
Simplify.
s=5 s=0
Add \frac{5}{2} to both sides of the equation.