Question 986583
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(2x\ +\ 8)\ =\ -4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ 8x\ +\ 4\ =\ 0]


The roots of:


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


are


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


If you first divide both sides of your equation by 2, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 1\ \ ], *[tex \LARGE b\ =\ 4\ \ ], and *[tex \LARGE c\ =\ 2].


Just plug in the numbers and do the arithmetic.  You can do your own calculator work, too.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \