Question 972857
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The value of *[tex \Large f(x)] when *[tex \Large f(x)\ =\ 0] is exactly *[tex \Large 0].  But I'm going to go out on a limb here and guess that you <i>really</i> meant to ask about the values of *[tex \Large x] such that *[tex \Large f(x)\ =\ 0].


Your little bit of ugliness doesn't factor, so your choices are to complete the square (if you are some sort of glutton for punishment) or use the quadratic formula.


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


Where 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ -9\ \ ], *[tex \LARGE b\ =\ 2\ \ ], and *[tex \LARGE c\ =\ 4]


Plug in the numbers and do the arithmetic to find your two values.


By the way, use a caret ("^") mark to indicate raising to a power in plain text.  -9x^2 + 2x + 4 means *[tex \Large -9x^2\ + 2x\ + 4] whereas -9x2 + 2x + 4 is just goobledygook.


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

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