Question 182228
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*[tex \LARGE \text{          }\math 4 \times 2 = 8]


*[tex \LARGE \text{          }\math -1 \times 2 = -2]


*[tex \LARGE \text{          }\math (4 \times 2) + (-1 \times 2) = 8 - 2 = 6]


Therefore:


*[tex \LARGE \text{          }\math 8x^2 + 6x -2 = 0\ \ \Rightarrow\ \ (4x - 1)(2x + 2) = 0\ \ \Rightarrow\ \ 4x - 1 = 0 \text{ or }2x + 2 = 0\cr\text{       }\ \ \Rightarrow\ \ 4x = 1\text{ or }2x = -2\ \ \Rightarrow\ \ x = \frac{1}{4} \text{ or }x = -1] 



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