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\frac{3}{2}a^{2}-a-\frac{1}{4}=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{-\left(-1\right)±\sqrt{1-4\times \frac{3}{2}\left(-\frac{1}{4}\right)}}{2\times \frac{3}{2}}
This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{3}{2} for a, -1 for b, and -\frac{1}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-\left(-1\right)±\sqrt{1-6\left(-\frac{1}{4}\right)}}{2\times \frac{3}{2}}
Multiply -4 times \frac{3}{2}.
a=\frac{-\left(-1\right)±\sqrt{1+\frac{3}{2}}}{2\times \frac{3}{2}}
Multiply -6 times -\frac{1}{4}.
a=\frac{-\left(-1\right)±\sqrt{\frac{5}{2}}}{2\times \frac{3}{2}}
Add 1 to \frac{3}{2}.
a=\frac{-\left(-1\right)±\frac{\sqrt{10}}{2}}{2\times \frac{3}{2}}
Take the square root of \frac{5}{2}.
a=\frac{1±\frac{\sqrt{10}}{2}}{2\times \frac{3}{2}}
The opposite of -1 is 1.
a=\frac{1±\frac{\sqrt{10}}{2}}{3}
Multiply 2 times \frac{3}{2}.
a=\frac{\frac{\sqrt{10}}{2}+1}{3}
Now solve the equation a=\frac{1±\frac{\sqrt{10}}{2}}{3} when ± is plus. Add 1 to \frac{\sqrt{10}}{2}.
a=\frac{\sqrt{10}}{6}+\frac{1}{3}
Divide 1+\frac{\sqrt{10}}{2} by 3.
a=\frac{-\frac{\sqrt{10}}{2}+1}{3}
Now solve the equation a=\frac{1±\frac{\sqrt{10}}{2}}{3} when ± is minus. Subtract \frac{\sqrt{10}}{2} from 1.
a=-\frac{\sqrt{10}}{6}+\frac{1}{3}
Divide 1-\frac{\sqrt{10}}{2} by 3.
a=\frac{\sqrt{10}}{6}+\frac{1}{3} a=-\frac{\sqrt{10}}{6}+\frac{1}{3}
The equation is now solved.
\frac{3}{2}a^{2}-a-\frac{1}{4}=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{3}{2}a^{2}-a-\frac{1}{4}-\left(-\frac{1}{4}\right)=-\left(-\frac{1}{4}\right)
Add \frac{1}{4} to both sides of the equation.
\frac{3}{2}a^{2}-a=-\left(-\frac{1}{4}\right)
Subtracting -\frac{1}{4} from itself leaves 0.
\frac{3}{2}a^{2}-a=\frac{1}{4}
Subtract -\frac{1}{4} from 0.
\frac{\frac{3}{2}a^{2}-a}{\frac{3}{2}}=\frac{\frac{1}{4}}{\frac{3}{2}}
Divide both sides of the equation by \frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.
a^{2}+\left(-\frac{1}{\frac{3}{2}}\right)a=\frac{\frac{1}{4}}{\frac{3}{2}}
Dividing by \frac{3}{2} undoes the multiplication by \frac{3}{2}.
a^{2}-\frac{2}{3}a=\frac{\frac{1}{4}}{\frac{3}{2}}
Divide -1 by \frac{3}{2} by multiplying -1 by the reciprocal of \frac{3}{2}.
a^{2}-\frac{2}{3}a=\frac{1}{6}
Divide \frac{1}{4} by \frac{3}{2} by multiplying \frac{1}{4} by the reciprocal of \frac{3}{2}.
a^{2}-\frac{2}{3}a+\left(-\frac{1}{3}\right)^{2}=\frac{1}{6}+\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.
a^{2}-\frac{2}{3}a+\frac{1}{9}=\frac{1}{6}+\frac{1}{9}
Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.
a^{2}-\frac{2}{3}a+\frac{1}{9}=\frac{5}{18}
Add \frac{1}{6} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(a-\frac{1}{3}\right)^{2}=\frac{5}{18}
Factor a^{2}-\frac{2}{3}a+\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(a-\frac{1}{3}\right)^{2}}=\sqrt{\frac{5}{18}}
Take the square root of both sides of the equation.
a-\frac{1}{3}=\frac{\sqrt{10}}{6} a-\frac{1}{3}=-\frac{\sqrt{10}}{6}
Simplify.
a=\frac{\sqrt{10}}{6}+\frac{1}{3} a=-\frac{\sqrt{10}}{6}+\frac{1}{3}
Add \frac{1}{3} to both sides of the equation.