Question 327255
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Put it into standard form and then use the quadratic formula.


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


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


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


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


There are no perfect square factors in the discriminant so simplifying is trivial.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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