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-\frac{1}{3}a^{2}+a+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.
a=\frac{-1±\sqrt{1^{2}-4\left(-\frac{1}{3}\right)\times 2}}{2\left(-\frac{1}{3}\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{3} for a, 1 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-1±\sqrt{1-4\left(-\frac{1}{3}\right)\times 2}}{2\left(-\frac{1}{3}\right)}
Square 1.
a=\frac{-1±\sqrt{1+\frac{4}{3}\times 2}}{2\left(-\frac{1}{3}\right)}
Multiply -4 times -\frac{1}{3}.
a=\frac{-1±\sqrt{1+\frac{8}{3}}}{2\left(-\frac{1}{3}\right)}
Multiply \frac{4}{3} times 2.
a=\frac{-1±\sqrt{\frac{11}{3}}}{2\left(-\frac{1}{3}\right)}
Add 1 to \frac{8}{3}.
a=\frac{-1±\frac{\sqrt{33}}{3}}{2\left(-\frac{1}{3}\right)}
Take the square root of \frac{11}{3}.
a=\frac{-1±\frac{\sqrt{33}}{3}}{-\frac{2}{3}}
Multiply 2 times -\frac{1}{3}.
a=\frac{\frac{\sqrt{33}}{3}-1}{-\frac{2}{3}}
Now solve the equation a=\frac{-1±\frac{\sqrt{33}}{3}}{-\frac{2}{3}} when ± is plus. Add -1 to \frac{\sqrt{33}}{3}.
a=\frac{3-\sqrt{33}}{2}
Divide -1+\frac{\sqrt{33}}{3} by -\frac{2}{3} by multiplying -1+\frac{\sqrt{33}}{3} by the reciprocal of -\frac{2}{3}.
a=\frac{-\frac{\sqrt{33}}{3}-1}{-\frac{2}{3}}
Now solve the equation a=\frac{-1±\frac{\sqrt{33}}{3}}{-\frac{2}{3}} when ± is minus. Subtract \frac{\sqrt{33}}{3} from -1.
a=\frac{\sqrt{33}+3}{2}
Divide -1-\frac{\sqrt{33}}{3} by -\frac{2}{3} by multiplying -1-\frac{\sqrt{33}}{3} by the reciprocal of -\frac{2}{3}.
a=\frac{3-\sqrt{33}}{2} a=\frac{\sqrt{33}+3}{2}
The equation is now solved.
-\frac{1}{3}a^{2}+a+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{1}{3}a^{2}+a+2-2=-2
Subtract 2 from both sides of the equation.
-\frac{1}{3}a^{2}+a=-2
Subtracting 2 from itself leaves 0.
\frac{-\frac{1}{3}a^{2}+a}{-\frac{1}{3}}=-\frac{2}{-\frac{1}{3}}
Multiply both sides by -3.
a^{2}+\frac{1}{-\frac{1}{3}}a=-\frac{2}{-\frac{1}{3}}
Dividing by -\frac{1}{3} undoes the multiplication by -\frac{1}{3}.
a^{2}-3a=-\frac{2}{-\frac{1}{3}}
Divide 1 by -\frac{1}{3} by multiplying 1 by the reciprocal of -\frac{1}{3}.
a^{2}-3a=6
Divide -2 by -\frac{1}{3} by multiplying -2 by the reciprocal of -\frac{1}{3}.
a^{2}-3a+\left(-\frac{3}{2}\right)^{2}=6+\left(-\frac{3}{2}\right)^{2}
Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}-3a+\frac{9}{4}=6+\frac{9}{4}
Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.
a^{2}-3a+\frac{9}{4}=\frac{33}{4}
Add 6 to \frac{9}{4}.
\left(a-\frac{3}{2}\right)^{2}=\frac{33}{4}
Factor a^{2}-3a+\frac{9}{4}. 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(a-\frac{3}{2}\right)^{2}}=\sqrt{\frac{33}{4}}
Take the square root of both sides of the equation.
a-\frac{3}{2}=\frac{\sqrt{33}}{2} a-\frac{3}{2}=-\frac{\sqrt{33}}{2}
Simplify.
a=\frac{\sqrt{33}+3}{2} a=\frac{3-\sqrt{33}}{2}
Add \frac{3}{2} to both sides of the equation.