Find $f$ by solving the initial-value problem. $f'(x) = 9x^2 + 6x - 2$; $f(2) = 5$ f(x) =
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To do this, we can use the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a where a = 9, b = 6, and c = -2. Plugging in these values, we get: x = (-6 ± sqrt(6^2 - 4(9)(-2))) / 2(9) x = (-6 ± sqrt(180)) / 18 x = (-6 ± 6sqrt(5)) / 18 x = (-1 ± sqrt(5)) / 3 So Show more…
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