Question 826588
 there are two consecutive natural numbers,such that twice the square of the first is 34 more than the square of the second. find the numbers 
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1st: x
2nd: x+1
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Equation:
2x^2 - (x+1)^2 = 34
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2x^2 - [x^2 + 2x + 1] = 34
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2x^2 - x^2 -2x -1 = 34
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x^2 -2x -35 = 0
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(x-7)(x+5) = 0
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Answers:
x = 7
x+1 = 8
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Cheers,
Stan H.
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