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7 square root of 3 end root minus 4 square root of 6 end root plus square root of 48 end root minus square root of 54

User Bharata
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7\sqrt3-4\sqrt6+√(48)-√(54)\\\\=7\sqrt3-4\sqrt6+√(16\cdot3)-√(9\cdot6)\\\\=7\sqrt3-4\sqrt6+√(16)\cdot\sqrt3-\sqrt9\cdot\sqrt6\\\\=7\sqrt3-4\sqrt6+4\sqrt3-3\sqrt6\\\\=\boxed{11\sqrt3-7\sqrt6}