Question 239452
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Step 1:  Put the equation into standard form, that is into


*[tex \Large \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ +\ c\ =\ 0]


form.


Your equation becomes:


*[tex \Large \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ 4x\ -\ 5\ =\ 0]


So, *[tex \Large a\ =\ 2], *[tex \Large b\ =\ 4], and *[tex \Large c\ =\ -5]


Then substitute the coefficients into the quadratic formula:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-(4) \pm sqrt{(4)^2 - 4(2)(-5)}}{2(2)} ]


You can do your own arithmetic.  To simplify the radical, perform the prime factorization of the radicand.  Remove pairs of prime factors from the radicand and put one of each pair outside the radicand.  Then reduce the fractions if possible.  Remember, this is a quadratic equation, so there are always two roots.


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