Das ist doch unsinnig. Da gibt es doch unendlich viele Lösungen
(2·x + 3)·([ ] + [ ]) = x·(2·x + 11) + [ ]
(2·x + 3)·([x] + [0]) = 2·x^2 + 11·x + [ ]
2·x^2 + 3·x = 2·x^2 + 11·x + [-8·x]
2·x^2 + 3·x = 2·x^2 + 11·x - 8·x
2·x^2 + 3·x = 2·x^2 + 3·x
Also z.B.
(2·x + 3)·([x] + [0]) = 2·x^2 + 11·x + [-8·x]