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3x^{2}-5x+\frac{1}{3}=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(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 3\times \frac{1}{3}}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -5 for b, and \frac{1}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-5\right)±\sqrt{25-4\times 3\times \frac{1}{3}}}{2\times 3}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25-12\times \frac{1}{3}}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-5\right)±\sqrt{25-4}}{2\times 3}
Multiply -12 times \frac{1}{3}.
x=\frac{-\left(-5\right)±\sqrt{21}}{2\times 3}
Add 25 to -4.
x=\frac{5±\sqrt{21}}{2\times 3}
The opposite of -5 is 5.
x=\frac{5±\sqrt{21}}{6}
Multiply 2 times 3.
x=\frac{\sqrt{21}+5}{6}
Now solve the equation x=\frac{5±\sqrt{21}}{6} when ± is plus. Add 5 to \sqrt{21}.
x=\frac{5-\sqrt{21}}{6}
Now solve the equation x=\frac{5±\sqrt{21}}{6} when ± is minus. Subtract \sqrt{21} from 5.
x=\frac{\sqrt{21}+5}{6} x=\frac{5-\sqrt{21}}{6}
The equation is now solved.
3x^{2}-5x+\frac{1}{3}=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.
3x^{2}-5x+\frac{1}{3}-\frac{1}{3}=-\frac{1}{3}
Subtract \frac{1}{3} from both sides of the equation.
3x^{2}-5x=-\frac{1}{3}
Subtracting \frac{1}{3} from itself leaves 0.
\frac{3x^{2}-5x}{3}=-\frac{\frac{1}{3}}{3}
Divide both sides by 3.
x^{2}-\frac{5}{3}x=-\frac{\frac{1}{3}}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-\frac{5}{3}x=-\frac{1}{9}
Divide -\frac{1}{3} by 3.
x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=-\frac{1}{9}+\left(-\frac{5}{6}\right)^{2}
Divide -\frac{5}{3}, the coefficient of the x term, by 2 to get -\frac{5}{6}. Then add the square of -\frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{5}{3}x+\frac{25}{36}=-\frac{1}{9}+\frac{25}{36}
Square -\frac{5}{6} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{5}{3}x+\frac{25}{36}=\frac{7}{12}
Add -\frac{1}{9} to \frac{25}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{5}{6}\right)^{2}=\frac{7}{12}
Factor x^{2}-\frac{5}{3}x+\frac{25}{36}. 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{5}{6}\right)^{2}}=\sqrt{\frac{7}{12}}
Take the square root of both sides of the equation.
x-\frac{5}{6}=\frac{\sqrt{21}}{6} x-\frac{5}{6}=-\frac{\sqrt{21}}{6}
Simplify.
x=\frac{\sqrt{21}+5}{6} x=\frac{5-\sqrt{21}}{6}
Add \frac{5}{6} to both sides of the equation.