Question 1083871
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(3Ay + B)*(y - C) = {{{3Ay^2 + By - 3ACy - BC}}} = {{{3Ay^2 + (B-3AC)y - BC}}}.


Since it is identical to {{{6y^2 - y - 51}}}, we have


    3A = 6  and hence A = 2;             (1)

    B - 3AC = -1,   or  B - 6C = -1;     (2)

    BC = 51.                             (3)


Thus you actually have these two equations to determine B and C:

B - 6C = -1                              (2)
BC = 51.                                 (3)


From (2), express B = 6C -1 and substitute it into (3). You will get

(6C-1)*C = 51.


{{{6C^2 - C - 51}}} = 0,

Factor left side

(C-3)*(2C+17) = 0.


Since you need C to be positive number (as the condition requires), you have only one possibility: C = 3.

Then B = 6C-1 = 6*3-1 = 17,

and now you have everything to calculate  {{{(AC)^2-B}}} = {{{(2*3)^2 - 17}}} = {{{6^2 - 17}}} = 19.
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