<PRE><B>Question 415944</B>
find three consecutive odd integers such that 4 times the product of the second and third is 12 greater than 20 times the sum of the first and second
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1st: 2x-1
2nd: 2x+1
3rd: 2x+3
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Equation:
4[(2x+1)(2x+3)] = 20[4x]
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4[4X^2+8x+3] = 80x
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4x^2+8x+3 = 20x
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4x^2-12x+3 = 0
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x = [12 +- sqrt(144 - 4*4*3)]/8
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x = [12 +- sqrt(96)]/8
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x = [12 +- 4sqrt(6)]/8
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x = (3/2) +- sqrt(6)
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Cheers,
Stan H.
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