Find the consecutive integers so that 4 times the square of the third decreased by 3 times the square of the first is 41 more than twice the square of the second

Remark
We have to assume that there are only 3 consecutive integers.

Givens
Let the smallest integer = x
Let the middle integer = x + 1
Let the largest integer = x + 2

Formulas
4(x + 2)^2 - 3x^2 = 41+ 2*(x + 1)^2 Expand the brackets.

Solution
4(x^2 + 4x + 4) - 3x^2 = 41 + 2(x^2 + 2x + 1)  Remove the brackets.
4x^2 + 16x + 16 - 3x^2 = 41 + 2x^2 + 4x + 2   Collect like terms on both sides
x^2 + 16x + 16 =2x^2 + 4x + 43                       Subtract the left side from the right

0 = 2x^2 - x^2 - 16x + 4x - 16 + 43                  Collect like terms.
0 = x^2 - 12x + 27                                            This factors
(x - 3) * (x - 9)  = 0

x - 3 = 0
x = 3        or
x - 9 = 0
x = 9

Therefore numbers can be 3,4,5 or 9,10,11

We need to check both.
First group check.
Group One
4*5^2 - 3*3^2 = 41 + 2*4^2
4*25 - 3*9 = 41 + 2*16
100 - 27 = 41 + 32
73 = 73
Group one is correct.

Group Two
4*11^2 - 3*9^2 = 41 + 2*10^2
484 - 243 = 41 + 200
241 = 241 and the second group checks as well. So thee are two sets of answers.

3,4,5 and 9,10,11 <<<<<< Answer.
Note: You should put this in math.

I  cant see the question.