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3x^{2}-\frac{4}{3}x-\frac{5}{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(-\frac{4}{3}\right)±\sqrt{\left(-\frac{4}{3}\right)^{2}-4\times 3\left(-\frac{5}{3}\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -\frac{4}{3} for b, and -\frac{5}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\frac{16}{9}-4\times 3\left(-\frac{5}{3}\right)}}{2\times 3}
Square -\frac{4}{3} by squaring both the numerator and the denominator of the fraction.
x=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\frac{16}{9}-12\left(-\frac{5}{3}\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\frac{16}{9}+20}}{2\times 3}
Multiply -12 times -\frac{5}{3}.
x=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\frac{196}{9}}}{2\times 3}
Add \frac{16}{9} to 20.
x=\frac{-\left(-\frac{4}{3}\right)±\frac{14}{3}}{2\times 3}
Take the square root of \frac{196}{9}.
x=\frac{\frac{4}{3}±\frac{14}{3}}{2\times 3}
The opposite of -\frac{4}{3} is \frac{4}{3}.
x=\frac{\frac{4}{3}±\frac{14}{3}}{6}
Multiply 2 times 3.
x=\frac{6}{6}
Now solve the equation x=\frac{\frac{4}{3}±\frac{14}{3}}{6} when ± is plus. Add \frac{4}{3} to \frac{14}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=1
Divide 6 by 6.
x=-\frac{\frac{10}{3}}{6}
Now solve the equation x=\frac{\frac{4}{3}±\frac{14}{3}}{6} when ± is minus. Subtract \frac{14}{3} from \frac{4}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
x=-\frac{5}{9}
Divide -\frac{10}{3} by 6.
x=1 x=-\frac{5}{9}
The equation is now solved.
3x^{2}-\frac{4}{3}x-\frac{5}{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}-\frac{4}{3}x-\frac{5}{3}-\left(-\frac{5}{3}\right)=-\left(-\frac{5}{3}\right)
Add \frac{5}{3} to both sides of the equation.
3x^{2}-\frac{4}{3}x=-\left(-\frac{5}{3}\right)
Subtracting -\frac{5}{3} from itself leaves 0.
3x^{2}-\frac{4}{3}x=\frac{5}{3}
Subtract -\frac{5}{3} from 0.
\frac{3x^{2}-\frac{4}{3}x}{3}=\frac{\frac{5}{3}}{3}
Divide both sides by 3.
x^{2}+\left(-\frac{\frac{4}{3}}{3}\right)x=\frac{\frac{5}{3}}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-\frac{4}{9}x=\frac{\frac{5}{3}}{3}
Divide -\frac{4}{3} by 3.
x^{2}-\frac{4}{9}x=\frac{5}{9}
Divide \frac{5}{3} by 3.
x^{2}-\frac{4}{9}x+\left(-\frac{2}{9}\right)^{2}=\frac{5}{9}+\left(-\frac{2}{9}\right)^{2}
Divide -\frac{4}{9}, the coefficient of the x term, by 2 to get -\frac{2}{9}. Then add the square of -\frac{2}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{4}{9}x+\frac{4}{81}=\frac{5}{9}+\frac{4}{81}
Square -\frac{2}{9} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{4}{9}x+\frac{4}{81}=\frac{49}{81}
Add \frac{5}{9} to \frac{4}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{2}{9}\right)^{2}=\frac{49}{81}
Factor x^{2}-\frac{4}{9}x+\frac{4}{81}. 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{2}{9}\right)^{2}}=\sqrt{\frac{49}{81}}
Take the square root of both sides of the equation.
x-\frac{2}{9}=\frac{7}{9} x-\frac{2}{9}=-\frac{7}{9}
Simplify.
x=1 x=-\frac{5}{9}
Add \frac{2}{9} to both sides of the equation.