<PRE><B>Question 602468</B>
Two consecutive positive integers such that the square of the second integer added to 4 times the first is equal to 28
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1st: x
2nd: x+1
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Equation:
(x+1)^2 + 4x = 28
x^2+2x+1 + 4x = 28
x^2+6x-27 = 0
Factor:
(x+9)(x-3) = 0
Positive solution:
1st: x = 3
2nd: x+1 = 4
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