Hi,

three consecutive integers , & 

 such that four times the square of the third, less than three times the square of the first, 

minus 41, is 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 

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