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\frac{3}{4}x^{2}+\frac{2}{3}x-\frac{1}{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{-\frac{2}{3}±\sqrt{\left(\frac{2}{3}\right)^{2}-4\times \frac{3}{4}\left(-\frac{1}{2}\right)}}{2\times \frac{3}{4}}
This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{3}{4} for a, \frac{2}{3} for b, and -\frac{1}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\frac{2}{3}±\sqrt{\frac{4}{9}-4\times \frac{3}{4}\left(-\frac{1}{2}\right)}}{2\times \frac{3}{4}}
Square \frac{2}{3} by squaring both the numerator and the denominator of the fraction.
x=\frac{-\frac{2}{3}±\sqrt{\frac{4}{9}-3\left(-\frac{1}{2}\right)}}{2\times \frac{3}{4}}
Multiply -4 times \frac{3}{4}.
x=\frac{-\frac{2}{3}±\sqrt{\frac{4}{9}+\frac{3}{2}}}{2\times \frac{3}{4}}
Multiply -3 times -\frac{1}{2}.
x=\frac{-\frac{2}{3}±\sqrt{\frac{35}{18}}}{2\times \frac{3}{4}}
Add \frac{4}{9} to \frac{3}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=\frac{-\frac{2}{3}±\frac{\sqrt{70}}{6}}{2\times \frac{3}{4}}
Take the square root of \frac{35}{18}.
x=\frac{-\frac{2}{3}±\frac{\sqrt{70}}{6}}{\frac{3}{2}}
Multiply 2 times \frac{3}{4}.
x=\frac{\frac{\sqrt{70}}{6}-\frac{2}{3}}{\frac{3}{2}}
Now solve the equation x=\frac{-\frac{2}{3}±\frac{\sqrt{70}}{6}}{\frac{3}{2}} when ± is plus. Add -\frac{2}{3} to \frac{\sqrt{70}}{6}.
x=\frac{\sqrt{70}-4}{9}
Divide -\frac{2}{3}+\frac{\sqrt{70}}{6} by \frac{3}{2} by multiplying -\frac{2}{3}+\frac{\sqrt{70}}{6} by the reciprocal of \frac{3}{2}.
x=\frac{-\frac{\sqrt{70}}{6}-\frac{2}{3}}{\frac{3}{2}}
Now solve the equation x=\frac{-\frac{2}{3}±\frac{\sqrt{70}}{6}}{\frac{3}{2}} when ± is minus. Subtract \frac{\sqrt{70}}{6} from -\frac{2}{3}.
x=\frac{-\sqrt{70}-4}{9}
Divide -\frac{2}{3}-\frac{\sqrt{70}}{6} by \frac{3}{2} by multiplying -\frac{2}{3}-\frac{\sqrt{70}}{6} by the reciprocal of \frac{3}{2}.
x=\frac{\sqrt{70}-4}{9} x=\frac{-\sqrt{70}-4}{9}
The equation is now solved.
\frac{3}{4}x^{2}+\frac{2}{3}x-\frac{1}{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{3}{4}x^{2}+\frac{2}{3}x-\frac{1}{2}-\left(-\frac{1}{2}\right)=-\left(-\frac{1}{2}\right)
Add \frac{1}{2} to both sides of the equation.
\frac{3}{4}x^{2}+\frac{2}{3}x=-\left(-\frac{1}{2}\right)
Subtracting -\frac{1}{2} from itself leaves 0.
\frac{3}{4}x^{2}+\frac{2}{3}x=\frac{1}{2}
Subtract -\frac{1}{2} from 0.
\frac{\frac{3}{4}x^{2}+\frac{2}{3}x}{\frac{3}{4}}=\frac{\frac{1}{2}}{\frac{3}{4}}
Divide both sides of the equation by \frac{3}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.
x^{2}+\frac{\frac{2}{3}}{\frac{3}{4}}x=\frac{\frac{1}{2}}{\frac{3}{4}}
Dividing by \frac{3}{4} undoes the multiplication by \frac{3}{4}.
x^{2}+\frac{8}{9}x=\frac{\frac{1}{2}}{\frac{3}{4}}
Divide \frac{2}{3} by \frac{3}{4} by multiplying \frac{2}{3} by the reciprocal of \frac{3}{4}.
x^{2}+\frac{8}{9}x=\frac{2}{3}
Divide \frac{1}{2} by \frac{3}{4} by multiplying \frac{1}{2} by the reciprocal of \frac{3}{4}.
x^{2}+\frac{8}{9}x+\left(\frac{4}{9}\right)^{2}=\frac{2}{3}+\left(\frac{4}{9}\right)^{2}
Divide \frac{8}{9}, the coefficient of the x term, by 2 to get \frac{4}{9}. Then add the square of \frac{4}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{8}{9}x+\frac{16}{81}=\frac{2}{3}+\frac{16}{81}
Square \frac{4}{9} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{8}{9}x+\frac{16}{81}=\frac{70}{81}
Add \frac{2}{3} to \frac{16}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{4}{9}\right)^{2}=\frac{70}{81}
Factor x^{2}+\frac{8}{9}x+\frac{16}{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{4}{9}\right)^{2}}=\sqrt{\frac{70}{81}}
Take the square root of both sides of the equation.
x+\frac{4}{9}=\frac{\sqrt{70}}{9} x+\frac{4}{9}=-\frac{\sqrt{70}}{9}
Simplify.
x=\frac{\sqrt{70}-4}{9} x=\frac{-\sqrt{70}-4}{9}
Subtract \frac{4}{9} from both sides of the equation.