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If $$ax^{2} + bx + 6$$ leaves the remainders 12 and 15 when divided by $$x - 3$$ and $$x + 4$$ respectively. then 7a + 14b =
$$\frac{9}{2}$$
$$\frac{17}{4}$$
$$\frac{27}{4}$$
$$\frac{11}{2}$$
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