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5x^{2}-x=25
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.
5x^{2}-x-25=25-25
Subtract 25 from both sides of the equation.
5x^{2}-x-25=0
Subtracting 25 from itself leaves 0.
x=\frac{-\left(-1\right)±\sqrt{1-4\times 5\left(-25\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -1 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-1\right)±\sqrt{1-20\left(-25\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-1\right)±\sqrt{1+500}}{2\times 5}
Multiply -20 times -25.
x=\frac{-\left(-1\right)±\sqrt{501}}{2\times 5}
Add 1 to 500.
x=\frac{1±\sqrt{501}}{2\times 5}
The opposite of -1 is 1.
x=\frac{1±\sqrt{501}}{10}
Multiply 2 times 5.
x=\frac{\sqrt{501}+1}{10}
Now solve the equation x=\frac{1±\sqrt{501}}{10} when ± is plus. Add 1 to \sqrt{501}.
x=\frac{1-\sqrt{501}}{10}
Now solve the equation x=\frac{1±\sqrt{501}}{10} when ± is minus. Subtract \sqrt{501} from 1.
x=\frac{\sqrt{501}+1}{10} x=\frac{1-\sqrt{501}}{10}
The equation is now solved.
5x^{2}-x=25
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{5x^{2}-x}{5}=\frac{25}{5}
Divide both sides by 5.
x^{2}-\frac{1}{5}x=\frac{25}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-\frac{1}{5}x=5
Divide 25 by 5.
x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=5+\left(-\frac{1}{10}\right)^{2}
Divide -\frac{1}{5}, the coefficient of the x term, by 2 to get -\frac{1}{10}. Then add the square of -\frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{1}{5}x+\frac{1}{100}=5+\frac{1}{100}
Square -\frac{1}{10} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{501}{100}
Add 5 to \frac{1}{100}.
\left(x-\frac{1}{10}\right)^{2}=\frac{501}{100}
Factor x^{2}-\frac{1}{5}x+\frac{1}{100}. 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-\frac{1}{10}\right)^{2}}=\sqrt{\frac{501}{100}}
Take the square root of both sides of the equation.
x-\frac{1}{10}=\frac{\sqrt{501}}{10} x-\frac{1}{10}=-\frac{\sqrt{501}}{10}
Simplify.
x=\frac{\sqrt{501}+1}{10} x=\frac{1-\sqrt{501}}{10}
Add \frac{1}{10} to both sides of the equation.