Question 657991
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Hi,
three consecutive integers {{{highlight(x)}}},{{{highlight(x+1)}}} & {{{highlight(x+2)}}}
 such that four times the square of the third, less than three times the square of the first, 
minus 41, <u>is</u> twice the square of the second
Question states***
 4(x+2)^2 - 3x^2 - 41 = 2(x+1)^2
 4(x^2 + 4x + 4) - 3x^2 - 41 = 2(x^2 + 2x+ 1)
 4x^2 + 16x + 16 - 3x^2 - 41 = 2x^2 + 4x + 2
                0 = x^2 - 12x + 27
                0 = (x-9)(x-3)
                  x = 9  0r x = 3
Integers are 9, 10, 11
0r  3,4,5