Question 179734
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*[tex \Large 3x^2=2x+5]


Step 1: Put the quadratic into standard form (*[tex \Large ax^2 + bx + c = 0]).  For this problem, add *[tex \Large -(2x + 5)] to both sides:


*[tex \Large 3x^2 - 2x - 5 = 0]


This quadratic factors:


*[tex \Large (x + 1)(3x - 5) = 0]


Using the Zero Product Rule:


*[tex \Large x + 1 = 0 \rightarrow x = -1]


or 


*[tex \Large 3x - 5 = 0 \rightarrow x = \frac {5}{3}]


Check your answers:



*[tex \Large 3(-1)^2=2(-1)+5 \rightarrow 3 = -2 + 5] True, answer checks.



*[tex \Large 3(\frac {5}{3})^2=2(\frac {5}{3})+5 \rightarrow \frac {25}{3} = \frac {10}{3} + \frac {15}{3}] True, answer checks.



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