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25x^{2}-\frac{9}{2}x-\frac{5}{2}=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{-\left(-\frac{9}{2}\right)±\sqrt{\left(-\frac{9}{2}\right)^{2}-4\times 25\left(-\frac{5}{2}\right)}}{2\times 25}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, -\frac{9}{2} for b, and -\frac{5}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-\frac{9}{2}\right)±\sqrt{\frac{81}{4}-4\times 25\left(-\frac{5}{2}\right)}}{2\times 25}
Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.
x=\frac{-\left(-\frac{9}{2}\right)±\sqrt{\frac{81}{4}-100\left(-\frac{5}{2}\right)}}{2\times 25}
Multiply -4 times 25.
x=\frac{-\left(-\frac{9}{2}\right)±\sqrt{\frac{81}{4}+250}}{2\times 25}
Multiply -100 times -\frac{5}{2}.
x=\frac{-\left(-\frac{9}{2}\right)±\sqrt{\frac{1081}{4}}}{2\times 25}
Add \frac{81}{4} to 250.
x=\frac{-\left(-\frac{9}{2}\right)±\frac{\sqrt{1081}}{2}}{2\times 25}
Take the square root of \frac{1081}{4}.
x=\frac{\frac{9}{2}±\frac{\sqrt{1081}}{2}}{2\times 25}
The opposite of -\frac{9}{2} is \frac{9}{2}.
x=\frac{\frac{9}{2}±\frac{\sqrt{1081}}{2}}{50}
Multiply 2 times 25.
x=\frac{\sqrt{1081}+9}{2\times 50}
Now solve the equation x=\frac{\frac{9}{2}±\frac{\sqrt{1081}}{2}}{50} when ± is plus. Add \frac{9}{2} to \frac{\sqrt{1081}}{2}.
x=\frac{\sqrt{1081}+9}{100}
Divide \frac{9+\sqrt{1081}}{2} by 50.
x=\frac{9-\sqrt{1081}}{2\times 50}
Now solve the equation x=\frac{\frac{9}{2}±\frac{\sqrt{1081}}{2}}{50} when ± is minus. Subtract \frac{\sqrt{1081}}{2} from \frac{9}{2}.
x=\frac{9-\sqrt{1081}}{100}
Divide \frac{9-\sqrt{1081}}{2} by 50.
x=\frac{\sqrt{1081}+9}{100} x=\frac{9-\sqrt{1081}}{100}
The equation is now solved.
25x^{2}-\frac{9}{2}x-\frac{5}{2}=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.
25x^{2}-\frac{9}{2}x-\frac{5}{2}-\left(-\frac{5}{2}\right)=-\left(-\frac{5}{2}\right)
Add \frac{5}{2} to both sides of the equation.
25x^{2}-\frac{9}{2}x=-\left(-\frac{5}{2}\right)
Subtracting -\frac{5}{2} from itself leaves 0.
25x^{2}-\frac{9}{2}x=\frac{5}{2}
Subtract -\frac{5}{2} from 0.
\frac{25x^{2}-\frac{9}{2}x}{25}=\frac{\frac{5}{2}}{25}
Divide both sides by 25.
x^{2}+\left(-\frac{\frac{9}{2}}{25}\right)x=\frac{\frac{5}{2}}{25}
Dividing by 25 undoes the multiplication by 25.
x^{2}-\frac{9}{50}x=\frac{\frac{5}{2}}{25}
Divide -\frac{9}{2} by 25.
x^{2}-\frac{9}{50}x=\frac{1}{10}
Divide \frac{5}{2} by 25.
x^{2}-\frac{9}{50}x+\left(-\frac{9}{100}\right)^{2}=\frac{1}{10}+\left(-\frac{9}{100}\right)^{2}
Divide -\frac{9}{50}, the coefficient of the x term, by 2 to get -\frac{9}{100}. Then add the square of -\frac{9}{100} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{9}{50}x+\frac{81}{10000}=\frac{1}{10}+\frac{81}{10000}
Square -\frac{9}{100} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{9}{50}x+\frac{81}{10000}=\frac{1081}{10000}
Add \frac{1}{10} to \frac{81}{10000} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{9}{100}\right)^{2}=\frac{1081}{10000}
Factor x^{2}-\frac{9}{50}x+\frac{81}{10000}. 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{9}{100}\right)^{2}}=\sqrt{\frac{1081}{10000}}
Take the square root of both sides of the equation.
x-\frac{9}{100}=\frac{\sqrt{1081}}{100} x-\frac{9}{100}=-\frac{\sqrt{1081}}{100}
Simplify.
x=\frac{\sqrt{1081}+9}{100} x=\frac{9-\sqrt{1081}}{100}
Add \frac{9}{100} to both sides of the equation.