[tex](x-3)^{2} \geq 2x(x-2)+6\\\\x^{2}-6x+9 \geq 2x^{2}-4x+6\\\\x^{2}-2x^{2}-6x+4x+9-6 \geq 0\\\\-x^{2}-2x+3 \geq 0\\\\\Delta = b^{2}-4ac = (-2)^{2}-4\cdot(-1)\cdot3 = 4 + 12 = 16\\\\\sqrt{\Delta} = \sqrt{16} = 4\\\\x_1 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-(-2)+4}{-2} = \frac{6}{-2} = -3\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-(-2)-4}{-2} = \frac{-2}{-2} = 1\\\\a < 0, \ ramiona \ paraboli \ skierowane \ do \ dolu\\\\x \in \ <-3;1>[/tex]
