Question 690231
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Technically, I can't answer your question. *[tex \LARGE 3x^2\ +\ 3x\ -\ 16] does not have any "solutions".  The <i>equation</i> *[tex \LARGE 3x^2\ +\ 3x\ -\ 16\ =\ 0] has two solutions, and the polynomial function *[tex \LARGE \rho(x)\ =\ 3x^2\ +\ 3x\ -\ 16] has two zeros.  But the way you asked it -- nope.


However, while I'm at it, I should mention that the quadratic equation *[tex \LARGE ax^2\ +\ bx\ +\ c\ =\ 0] has two roots, namely:


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


and


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


Adding these two roots you get:


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


Presuming you meant to state your problem correctly, you should be able to glean the required answer now. 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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