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b\left(b-5\right)=0
Factor out b.
b=0 b=5
To find equation solutions, solve b=0 and b-5=0.
b^{2}-5b=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.
b=\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}.
b=\frac{-\left(-5\right)±5}{2}
Take the square root of \left(-5\right)^{2}.
b=\frac{5±5}{2}
The opposite of -5 is 5.
b=\frac{10}{2}
Now solve the equation b=\frac{5±5}{2} when ± is plus. Add 5 to 5.
b=5
Divide 10 by 2.
b=\frac{0}{2}
Now solve the equation b=\frac{5±5}{2} when ± is minus. Subtract 5 from 5.
b=0
Divide 0 by 2.
b=5 b=0
The equation is now solved.
b^{2}-5b=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.
b^{2}-5b+\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.
b^{2}-5b+\frac{25}{4}=\frac{25}{4}
Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.
\left(b-\frac{5}{2}\right)^{2}=\frac{25}{4}
Factor b^{2}-5b+\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(b-\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}
Take the square root of both sides of the equation.
b-\frac{5}{2}=\frac{5}{2} b-\frac{5}{2}=-\frac{5}{2}
Simplify.
b=5 b=0
Add \frac{5}{2} to both sides of the equation.