Question 181480
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The best approach to both of these is to get rid of the fractional coefficients.  For instance,


*[tex \LARGE \text{          }\math -6\frac{2}{3}x^2 + 1\frac{1}{3}x = \frac{2}{3}]


Start by converting the mixed fractions to improper fractions:


*[tex \LARGE \text{          }\math -\frac{20}{3}x^2 + \frac{4}{3}x = \frac{2}{3}]


Multiply both sides by -3:


*[tex \LARGE \text{          }\math 20x^2 - 4x = -2]


Add 2 to both sides:


*[tex \LARGE \text{          }\math 20x^2 - 4x + 2 = 0]


Multiply both sides by *[tex \Large \frac{1}{2}]


*[tex \LARGE \text{          }\math 10x^2 - 2x + 1 = 0]


Now you have an equivalent equation with integer factors that can be easily solved with the quadratic formula.


Follow a similar process for your second problem.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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