KCC's Quizzes AQQ262 about conical bottle height

Volume of a right cone is  (1/3)  pi R^2 H.   Given r = p h  at  an elevation h, ( with p = R/H)   
                        we deal with  V = K  h^3       with K = (1/3) pi  p  
Volume of the water at the left = K H^3  - K*8^3   =  volume of water at the right = K (H-2) ^3
Dividing by the constant K which is not zero, and evaluation the polynomials, we get:  4H^2 - 8H -504  = 0  ;  note:  504 =  8^3 - 2^3

We get  H  = 1 +  sqrt(127) ,  (the other root is negative)  and greater than 8,  OK  (numerially:  12.269... )

The volume of water is dependant of the radius of the circle at the bottom of the cone, but any R gives the H valeur here up and thus, R is undecided.

Volume of a right cone is  (1/3)  pi R^2 H.   Given r = p h  at  an elevation h, ( with p = R/H)   
                        we deal with  V = K  h^3       with K = (1/3) pi  p  
Volume of the water at the left = K H^3  - K*8^3   =  volume of water at the right = K (H-2) ^3
Dividing by the constant K which is not zero, and evaluation the polynomials, we get:  4H^2 - 8H -504  = 0  ;  note:  504 =  8^3 - 2^3

We get  H  = 1 +  sqrt(127) ,  (the other root is negative)  and greater than 8,  OK  (numerially:  12.269... )

The volume of water is dependant of the radius of the circle at the bottom of the cone, but any R gives the H valeur here up and thus, R is undecided.

Thanks vanderghast for your prompt feedback! Your development is correct but I think you made a typo somewhere before ending to the polynomial 4H^2 - 8H -504  = 0

504 seems to be incorrect...

May be you can recheck?

Got a tough time to find back my original computation (the sheet of paper is far in the recycling bin) but indeed, it seems that I tripped on the development of (H-2)^3. I now get 6H^2 -12H -504 = 0, leading to H= 10.2195... 

YES! You get it now vanderghast , congratulations!