<PRE><B>Question 413304</B>
find three consecutive integers such that the square of twice the first is
 30 more than three times the product of the second and third?
:
Three consecutive integers, x, (x+1). (x+2)
:
(2x)^2 = 3(x+1)*(x+2) + 30
:
square 2x and FOIL
4x^2 = 3(x^2 + 3x + 2) + 30

4x^2 = 3x^2 + 9x + 6 + 30
:
Combine like terms on the left
4x^2 - 3x^2 - 9x - 36 = 0
:
x^2 - 9x - 36 = 9
Factors to 
(x - 12)(x + 3) = 0
Two solutions
x = 12
and 
x = -3
:
12, 13, 14 are our 3 integers
or
-3, -2, -1, could also be the 3 integers
:
:
Check both in the statement
(2*12)^2 = 3(13*14) + 30
576 = 546 + 30; confirms our solution of x=12
and
(2*-3)^2 = 3(-2 * -1) + 30
36 = 6 + 30; confirms the x = -3 solution
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