$$33\cdot\left(\frac{2}{11}\right)^2-28\cdot\frac{2}{11} = 33\cdot\frac{4}{121}-28\cdot\frac{2}{11} = \frac{33\cdot4}{121}-\frac{28\cdot2}{11} = \frac{3\cdot4}{11}-\frac{28\cdot2}{11} = \frac{12}{11}-\frac{56}{11} = \frac{12-56}{11} = \frac{-44}{11} = -4$$
Ответ: -4