Question 584075
<font face="Times New Roman" size="+2">


You cannot actually solve the equation as you have written it. Since X and x are two different things, you actually have a single equation in two variables.  However, presuming a typo, proceed thus: 


Step 1:  Put your equation into standard form, i.e. add the opposite of the constant term to both sides so that everything is in the LHS and the RHS contains zero, thus:


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


For your problem, 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 1\ \ ], *[tex \LARGE b\ =\ -5\ \ ], and *[tex \LARGE c\ =\ -14]


The quadratic formula is:


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


Just plug in the values for *[tex \LARGE a], *[tex \LARGE b], and *[tex \LARGE c] and do the arithmetic.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>