[tex](x-1)^{2}\leq \frac{3}{2}\\\\x^{2}-2x+1\leq \frac{3}{2} \ \ |\cdot2\\\\2x^{2}-4x+2 \leq 3\\\\2x^{2}-4x+2-3 \leq 0\\\\2x^{2}-4x-1 \leq 0\\\\a = 2, \ b = -4, \ c = -1\\\\\Delta = b^{2}-4ac = (-4)^{2}-4\cdot2\cdot(-1) = 16+8 = 24\\\\\sqrt{\Delta} = \sqrt{24} = \sqrt{4\cdot6} = 2\sqrt{6}\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-(-4)-2\sqrt{6}}{2\cdot2} = \frac{4-2\sqrt{6}}{4}\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{4+2\sqrt{6}}{4}[/tex]
a > 0, to parabola zwrócona jest ramionami do dóry, wówczas wartości ≤ 0 znajdują się w przedziale:
[tex]\boxed{x \in \langle\frac{4-2\sqrt{6}}{4}; \frac{4+2\sqrt{6}}{4}\rangle}[/tex]
