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\frac{2}{3}x^{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(-1\right)±\sqrt{1-4\times \frac{2}{3}\times \frac{5}{2}}}{2\times \frac{2}{3}}
This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{2}{3} for a, -1 for b, and \frac{5}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-1\right)±\sqrt{1-\frac{8}{3}\times \frac{5}{2}}}{2\times \frac{2}{3}}
Multiply -4 times \frac{2}{3}.
x=\frac{-\left(-1\right)±\sqrt{1-\frac{20}{3}}}{2\times \frac{2}{3}}
Multiply -\frac{8}{3} times \frac{5}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=\frac{-\left(-1\right)±\sqrt{-\frac{17}{3}}}{2\times \frac{2}{3}}
Add 1 to -\frac{20}{3}.
x=\frac{-\left(-1\right)±\frac{\sqrt{51}i}{3}}{2\times \frac{2}{3}}
Take the square root of -\frac{17}{3}.
x=\frac{1±\frac{\sqrt{51}i}{3}}{2\times \frac{2}{3}}
The opposite of -1 is 1.
x=\frac{1±\frac{\sqrt{51}i}{3}}{\frac{4}{3}}
Multiply 2 times \frac{2}{3}.
x=\frac{\frac{\sqrt{51}i}{3}+1}{\frac{4}{3}}
Now solve the equation x=\frac{1±\frac{\sqrt{51}i}{3}}{\frac{4}{3}} when ± is plus. Add 1 to \frac{i\sqrt{51}}{3}.
x=\frac{3+\sqrt{51}i}{4}
Divide 1+\frac{i\sqrt{51}}{3} by \frac{4}{3} by multiplying 1+\frac{i\sqrt{51}}{3} by the reciprocal of \frac{4}{3}.
x=\frac{-\frac{\sqrt{51}i}{3}+1}{\frac{4}{3}}
Now solve the equation x=\frac{1±\frac{\sqrt{51}i}{3}}{\frac{4}{3}} when ± is minus. Subtract \frac{i\sqrt{51}}{3} from 1.
x=\frac{-\sqrt{51}i+3}{4}
Divide 1-\frac{i\sqrt{51}}{3} by \frac{4}{3} by multiplying 1-\frac{i\sqrt{51}}{3} by the reciprocal of \frac{4}{3}.
x=\frac{3+\sqrt{51}i}{4} x=\frac{-\sqrt{51}i+3}{4}
The equation is now solved.
\frac{2}{3}x^{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.
\frac{2}{3}x^{2}-x+\frac{5}{2}-\frac{5}{2}=-\frac{5}{2}
Subtract \frac{5}{2} from both sides of the equation.
\frac{2}{3}x^{2}-x=-\frac{5}{2}
Subtracting \frac{5}{2} from itself leaves 0.
\frac{\frac{2}{3}x^{2}-x}{\frac{2}{3}}=-\frac{\frac{5}{2}}{\frac{2}{3}}
Divide both sides of the equation by \frac{2}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.
x^{2}+\left(-\frac{1}{\frac{2}{3}}\right)x=-\frac{\frac{5}{2}}{\frac{2}{3}}
Dividing by \frac{2}{3} undoes the multiplication by \frac{2}{3}.
x^{2}-\frac{3}{2}x=-\frac{\frac{5}{2}}{\frac{2}{3}}
Divide -1 by \frac{2}{3} by multiplying -1 by the reciprocal of \frac{2}{3}.
x^{2}-\frac{3}{2}x=-\frac{15}{4}
Divide -\frac{5}{2} by \frac{2}{3} by multiplying -\frac{5}{2} by the reciprocal of \frac{2}{3}.
x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=-\frac{15}{4}+\left(-\frac{3}{4}\right)^{2}
Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{15}{4}+\frac{9}{16}
Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{51}{16}
Add -\frac{15}{4} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{3}{4}\right)^{2}=-\frac{51}{16}
Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. 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}{4}\right)^{2}}=\sqrt{-\frac{51}{16}}
Take the square root of both sides of the equation.
x-\frac{3}{4}=\frac{\sqrt{51}i}{4} x-\frac{3}{4}=-\frac{\sqrt{51}i}{4}
Simplify.
x=\frac{3+\sqrt{51}i}{4} x=\frac{-\sqrt{51}i+3}{4}
Add \frac{3}{4} to both sides of the equation.