Question 437475
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First put your equation into standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6x^2\ +\ 12x\ =\ -5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6x^2\ +\ 12x\ +\ 5\ =\ 0]


The quadratic formula


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


gives the solutions to


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


First determine the coefficient values by inspection:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 6\ \ ], *[tex \LARGE b\ =\ 12\ \ ], and *[tex \LARGE c\ =\ 5]


Then substitute the values into the quadratic equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-12\ \pm\ \sqrt{(12)^2\ -\ 4(6)(5)}}{2(6)}]


All that is left is a little arithmetic.  In the event the radicand is not a perfect square, leave the answer in simplified radical form for the exact answer.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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