75 = \frac { 1 } { 2 } b ( b - 5
Solve for b
b=-10
b=15
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75=\frac{1}{2}bb+\frac{1}{2}b\left(-5\right)
Use the distributive property to multiply \frac{1}{2}b by b-5.
75=\frac{1}{2}b^{2}+\frac{1}{2}b\left(-5\right)
Multiply b and b to get b^{2}.
75=\frac{1}{2}b^{2}+\frac{-5}{2}b
Multiply \frac{1}{2} and -5 to get \frac{-5}{2}.
75=\frac{1}{2}b^{2}-\frac{5}{2}b
Fraction \frac{-5}{2} can be rewritten as -\frac{5}{2} by extracting the negative sign.
\frac{1}{2}b^{2}-\frac{5}{2}b=75
Swap sides so that all variable terms are on the left hand side.
\frac{1}{2}b^{2}-\frac{5}{2}b-75=0
Subtract 75 from both sides.
b=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\left(-\frac{5}{2}\right)^{2}-4\times \frac{1}{2}\left(-75\right)}}{2\times \frac{1}{2}}
This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, -\frac{5}{2} for b, and -75 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
b=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25}{4}-4\times \frac{1}{2}\left(-75\right)}}{2\times \frac{1}{2}}
Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.
b=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25}{4}-2\left(-75\right)}}{2\times \frac{1}{2}}
Multiply -4 times \frac{1}{2}.
b=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25}{4}+150}}{2\times \frac{1}{2}}
Multiply -2 times -75.
b=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{625}{4}}}{2\times \frac{1}{2}}
Add \frac{25}{4} to 150.
b=\frac{-\left(-\frac{5}{2}\right)±\frac{25}{2}}{2\times \frac{1}{2}}
Take the square root of \frac{625}{4}.
b=\frac{\frac{5}{2}±\frac{25}{2}}{2\times \frac{1}{2}}
The opposite of -\frac{5}{2} is \frac{5}{2}.
b=\frac{\frac{5}{2}±\frac{25}{2}}{1}
Multiply 2 times \frac{1}{2}.
b=\frac{15}{1}
Now solve the equation b=\frac{\frac{5}{2}±\frac{25}{2}}{1} when ± is plus. Add \frac{5}{2} to \frac{25}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
b=15
Divide 15 by 1.
b=-\frac{10}{1}
Now solve the equation b=\frac{\frac{5}{2}±\frac{25}{2}}{1} when ± is minus. Subtract \frac{25}{2} from \frac{5}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
b=-10
Divide -10 by 1.
b=15 b=-10
The equation is now solved.
75=\frac{1}{2}bb+\frac{1}{2}b\left(-5\right)
Use the distributive property to multiply \frac{1}{2}b by b-5.
75=\frac{1}{2}b^{2}+\frac{1}{2}b\left(-5\right)
Multiply b and b to get b^{2}.
75=\frac{1}{2}b^{2}+\frac{-5}{2}b
Multiply \frac{1}{2} and -5 to get \frac{-5}{2}.
75=\frac{1}{2}b^{2}-\frac{5}{2}b
Fraction \frac{-5}{2} can be rewritten as -\frac{5}{2} by extracting the negative sign.
\frac{1}{2}b^{2}-\frac{5}{2}b=75
Swap sides so that all variable terms are on the left hand side.
\frac{\frac{1}{2}b^{2}-\frac{5}{2}b}{\frac{1}{2}}=\frac{75}{\frac{1}{2}}
Multiply both sides by 2.
b^{2}+\left(-\frac{\frac{5}{2}}{\frac{1}{2}}\right)b=\frac{75}{\frac{1}{2}}
Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.
b^{2}-5b=\frac{75}{\frac{1}{2}}
Divide -\frac{5}{2} by \frac{1}{2} by multiplying -\frac{5}{2} by the reciprocal of \frac{1}{2}.
b^{2}-5b=150
Divide 75 by \frac{1}{2} by multiplying 75 by the reciprocal of \frac{1}{2}.
b^{2}-5b+\left(-\frac{5}{2}\right)^{2}=150+\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}=150+\frac{25}{4}
Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.
b^{2}-5b+\frac{25}{4}=\frac{625}{4}
Add 150 to \frac{25}{4}.
\left(b-\frac{5}{2}\right)^{2}=\frac{625}{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{625}{4}}
Take the square root of both sides of the equation.
b-\frac{5}{2}=\frac{25}{2} b-\frac{5}{2}=-\frac{25}{2}
Simplify.
b=15 b=-10
Add \frac{5}{2} to both sides of the equation.
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Limits
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