Solve for x
x=5\sqrt{6}+30\approx 42.247448714
x=30-5\sqrt{6}\approx 17.752551286
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40x-\frac{1}{2}x^{2}-20x=-\frac{1}{6}x^{2}+250
Subtract 20x from both sides.
20x-\frac{1}{2}x^{2}=-\frac{1}{6}x^{2}+250
Combine 40x and -20x to get 20x.
20x-\frac{1}{2}x^{2}+\frac{1}{6}x^{2}=250
Add \frac{1}{6}x^{2} to both sides.
20x-\frac{1}{3}x^{2}=250
Combine -\frac{1}{2}x^{2} and \frac{1}{6}x^{2} to get -\frac{1}{3}x^{2}.
20x-\frac{1}{3}x^{2}-250=0
Subtract 250 from both sides.
-\frac{1}{3}x^{2}+20x-250=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}-4\left(-\frac{1}{3}\right)\left(-250\right)}}{2\left(-\frac{1}{3}\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{3} for a, 20 for b, and -250 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-20±\sqrt{400-4\left(-\frac{1}{3}\right)\left(-250\right)}}{2\left(-\frac{1}{3}\right)}
Square 20.
x=\frac{-20±\sqrt{400+\frac{4}{3}\left(-250\right)}}{2\left(-\frac{1}{3}\right)}
Multiply -4 times -\frac{1}{3}.
x=\frac{-20±\sqrt{400-\frac{1000}{3}}}{2\left(-\frac{1}{3}\right)}
Multiply \frac{4}{3} times -250.
x=\frac{-20±\sqrt{\frac{200}{3}}}{2\left(-\frac{1}{3}\right)}
Add 400 to -\frac{1000}{3}.
x=\frac{-20±\frac{10\sqrt{6}}{3}}{2\left(-\frac{1}{3}\right)}
Take the square root of \frac{200}{3}.
x=\frac{-20±\frac{10\sqrt{6}}{3}}{-\frac{2}{3}}
Multiply 2 times -\frac{1}{3}.
x=\frac{\frac{10\sqrt{6}}{3}-20}{-\frac{2}{3}}
Now solve the equation x=\frac{-20±\frac{10\sqrt{6}}{3}}{-\frac{2}{3}} when ± is plus. Add -20 to \frac{10\sqrt{6}}{3}.
x=30-5\sqrt{6}
Divide -20+\frac{10\sqrt{6}}{3} by -\frac{2}{3} by multiplying -20+\frac{10\sqrt{6}}{3} by the reciprocal of -\frac{2}{3}.
x=\frac{-\frac{10\sqrt{6}}{3}-20}{-\frac{2}{3}}
Now solve the equation x=\frac{-20±\frac{10\sqrt{6}}{3}}{-\frac{2}{3}} when ± is minus. Subtract \frac{10\sqrt{6}}{3} from -20.
x=5\sqrt{6}+30
Divide -20-\frac{10\sqrt{6}}{3} by -\frac{2}{3} by multiplying -20-\frac{10\sqrt{6}}{3} by the reciprocal of -\frac{2}{3}.
x=30-5\sqrt{6} x=5\sqrt{6}+30
The equation is now solved.
40x-\frac{1}{2}x^{2}-20x=-\frac{1}{6}x^{2}+250
Subtract 20x from both sides.
20x-\frac{1}{2}x^{2}=-\frac{1}{6}x^{2}+250
Combine 40x and -20x to get 20x.
20x-\frac{1}{2}x^{2}+\frac{1}{6}x^{2}=250
Add \frac{1}{6}x^{2} to both sides.
20x-\frac{1}{3}x^{2}=250
Combine -\frac{1}{2}x^{2} and \frac{1}{6}x^{2} to get -\frac{1}{3}x^{2}.
-\frac{1}{3}x^{2}+20x=250
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{-\frac{1}{3}x^{2}+20x}{-\frac{1}{3}}=\frac{250}{-\frac{1}{3}}
Multiply both sides by -3.
x^{2}+\frac{20}{-\frac{1}{3}}x=\frac{250}{-\frac{1}{3}}
Dividing by -\frac{1}{3} undoes the multiplication by -\frac{1}{3}.
x^{2}-60x=\frac{250}{-\frac{1}{3}}
Divide 20 by -\frac{1}{3} by multiplying 20 by the reciprocal of -\frac{1}{3}.
x^{2}-60x=-750
Divide 250 by -\frac{1}{3} by multiplying 250 by the reciprocal of -\frac{1}{3}.
x^{2}-60x+\left(-30\right)^{2}=-750+\left(-30\right)^{2}
Divide -60, the coefficient of the x term, by 2 to get -30. Then add the square of -30 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-60x+900=-750+900
Square -30.
x^{2}-60x+900=150
Add -750 to 900.
\left(x-30\right)^{2}=150
Factor x^{2}-60x+900. 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-30\right)^{2}}=\sqrt{150}
Take the square root of both sides of the equation.
x-30=5\sqrt{6} x-30=-5\sqrt{6}
Simplify.
x=5\sqrt{6}+30 x=30-5\sqrt{6}
Add 30 to both sides of the equation.
Examples
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Linear equation
y = 3x + 4
Arithmetic
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Matrix
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Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
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Limits
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