Question 330422
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The given equation can be solved using the quadratic formula.  I have no idea what a quadractic formula is.


For a quadratic equation in standard form, namely:


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


the solutions are:


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


So, put the equation into standard form:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 3x\ -\ 7\ =\ 0]


For your equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 1\ \ ], *[tex \LARGE b\ =\ 3\ \ ], and *[tex \LARGE c\ =\ -7]


Just plug the values into the formula and do the arithmetic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-(3)\ \pm\ \sqrt{(3)^2\ -\ 4(1)(-7)}}{2(1)}]



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