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x^{2}-\frac{2}{3}x+\frac{9}{5}=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{2}{3}\right)±\sqrt{\left(-\frac{2}{3}\right)^{2}-4\times \frac{9}{5}}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -\frac{2}{3} for b, and \frac{9}{5} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-\frac{2}{3}\right)±\sqrt{\frac{4}{9}-4\times \frac{9}{5}}}{2}
Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.
x=\frac{-\left(-\frac{2}{3}\right)±\sqrt{\frac{4}{9}-\frac{36}{5}}}{2}
Multiply -4 times \frac{9}{5}.
x=\frac{-\left(-\frac{2}{3}\right)±\sqrt{-\frac{304}{45}}}{2}
Add \frac{4}{9} to -\frac{36}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=\frac{-\left(-\frac{2}{3}\right)±\frac{4\sqrt{95}i}{15}}{2}
Take the square root of -\frac{304}{45}.
x=\frac{\frac{2}{3}±\frac{4\sqrt{95}i}{15}}{2}
The opposite of -\frac{2}{3} is \frac{2}{3}.
x=\frac{\frac{4\sqrt{95}i}{15}+\frac{2}{3}}{2}
Now solve the equation x=\frac{\frac{2}{3}±\frac{4\sqrt{95}i}{15}}{2} when ± is plus. Add \frac{2}{3} to \frac{4i\sqrt{95}}{15}.
x=\frac{2\sqrt{95}i}{15}+\frac{1}{3}
Divide \frac{2}{3}+\frac{4i\sqrt{95}}{15} by 2.
x=\frac{-\frac{4\sqrt{95}i}{15}+\frac{2}{3}}{2}
Now solve the equation x=\frac{\frac{2}{3}±\frac{4\sqrt{95}i}{15}}{2} when ± is minus. Subtract \frac{4i\sqrt{95}}{15} from \frac{2}{3}.
x=-\frac{2\sqrt{95}i}{15}+\frac{1}{3}
Divide \frac{2}{3}-\frac{4i\sqrt{95}}{15} by 2.
x=\frac{2\sqrt{95}i}{15}+\frac{1}{3} x=-\frac{2\sqrt{95}i}{15}+\frac{1}{3}
The equation is now solved.
x^{2}-\frac{2}{3}x+\frac{9}{5}=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.
x^{2}-\frac{2}{3}x+\frac{9}{5}-\frac{9}{5}=-\frac{9}{5}
Subtract \frac{9}{5} from both sides of the equation.
x^{2}-\frac{2}{3}x=-\frac{9}{5}
Subtracting \frac{9}{5} from itself leaves 0.
x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=-\frac{9}{5}+\left(-\frac{1}{3}\right)^{2}
Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{9}{5}+\frac{1}{9}
Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{76}{45}
Add -\frac{9}{5} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{3}\right)^{2}=-\frac{76}{45}
Factor x^{2}-\frac{2}{3}x+\frac{1}{9}. 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}{3}\right)^{2}}=\sqrt{-\frac{76}{45}}
Take the square root of both sides of the equation.
x-\frac{1}{3}=\frac{2\sqrt{95}i}{15} x-\frac{1}{3}=-\frac{2\sqrt{95}i}{15}
Simplify.
x=\frac{2\sqrt{95}i}{15}+\frac{1}{3} x=-\frac{2\sqrt{95}i}{15}+\frac{1}{3}
Add \frac{1}{3} to both sides of the equation.