Question 234676
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^2\ +\ 2x\ -\ 4\ =\ 0]


So, *[tex \Large a\ =\ 3], *[tex \Large b\ =\ 2], and *[tex \Large c\ = -4]


The quadratic formula is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-b \pm sqrt{b^2 - 4ac}}{2a} ]


Just plug in the coefficient values and do the arithmetic.  Simplify the radical by factoring out one each of any pairs of prime factors of the radicand.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-(2) \pm sqrt{(2)^2 - 4(3)(-4)}}{2(3)} ]


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