Since the remaining water changes quadractically with time,

litres remaining can be written as ` L = at^2 + bt + c` where `t` is time since you pulled the plug.

At `t=1`  `at^2 + bt + c = a + b + c= 38.4`

At `t=2`  `at^2 + bt + c = 4a + 2b + c = 21.6`

At `t=3`  `at^2 + bt + c = 9a + 3b + c = 9.6`

Reduce to Echelon form by subtracting row 1) from rows 2) and 3)

1) 1  1  1   38.4                  1)  1   1   1   38.4

2) 4  2  1   21.6           2)-4*1)  0  -2  -3   -132

3) 9  3  1   9.6            3)-9*1)   0  -6  -8   -336

Then multiplying row 2) by -1/2

1) 1   1   1     38.4

2) 0   1   3/2  66

3) 0  -6  -8    -336

Then adding row 2) to row 3)

         1) 1   1    1     38.4

         2) 0   1    3/2   66

  3)+6*2) 0   0   1      60

` `

From equation 3) `c = 60`

substituting into equation 2)  `b + 3/2(60) = 66 implies b = -24`

substituting into equation 1) `a -24 + 60 = 38.4 implies a = 2.4`

Therefore the equation for remaining litres after t minutes is given by

L = 2.4^2 -24t +60

Check the solution at 1,2 and 3 minutes to confirm