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5x^{2}-3x+1=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(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 5}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -3 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-3\right)±\sqrt{9-4\times 5}}{2\times 5}
Square -3.
x=\frac{-\left(-3\right)±\sqrt{9-20}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-3\right)±\sqrt{-11}}{2\times 5}
Add 9 to -20.
x=\frac{-\left(-3\right)±\sqrt{11}i}{2\times 5}
Take the square root of -11.
x=\frac{3±\sqrt{11}i}{2\times 5}
The opposite of -3 is 3.
x=\frac{3±\sqrt{11}i}{10}
Multiply 2 times 5.
x=\frac{3+\sqrt{11}i}{10}
Now solve the equation x=\frac{3±\sqrt{11}i}{10} when ± is plus. Add 3 to i\sqrt{11}.
x=\frac{-\sqrt{11}i+3}{10}
Now solve the equation x=\frac{3±\sqrt{11}i}{10} when ± is minus. Subtract i\sqrt{11} from 3.
x=\frac{3+\sqrt{11}i}{10} x=\frac{-\sqrt{11}i+3}{10}
The equation is now solved.
5x^{2}-3x+1=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.
5x^{2}-3x+1-1=-1
Subtract 1 from both sides of the equation.
5x^{2}-3x=-1
Subtracting 1 from itself leaves 0.
\frac{5x^{2}-3x}{5}=-\frac{1}{5}
Divide both sides by 5.
x^{2}-\frac{3}{5}x=-\frac{1}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-\frac{3}{5}x+\left(-\frac{3}{10}\right)^{2}=-\frac{1}{5}+\left(-\frac{3}{10}\right)^{2}
Divide -\frac{3}{5}, the coefficient of the x term, by 2 to get -\frac{3}{10}. Then add the square of -\frac{3}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{3}{5}x+\frac{9}{100}=-\frac{1}{5}+\frac{9}{100}
Square -\frac{3}{10} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{3}{5}x+\frac{9}{100}=-\frac{11}{100}
Add -\frac{1}{5} to \frac{9}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{3}{10}\right)^{2}=-\frac{11}{100}
Factor x^{2}-\frac{3}{5}x+\frac{9}{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{3}{10}\right)^{2}}=\sqrt{-\frac{11}{100}}
Take the square root of both sides of the equation.
x-\frac{3}{10}=\frac{\sqrt{11}i}{10} x-\frac{3}{10}=-\frac{\sqrt{11}i}{10}
Simplify.
x=\frac{3+\sqrt{11}i}{10} x=\frac{-\sqrt{11}i+3}{10}
Add \frac{3}{10} to both sides of the equation.