Найдём корни квадратного трёхчлена через дискриминант:

[tex]2 {x}^{2} - 3x + 1 = 0 \\ a =2 \\ b = - 3 \\ c =1 \\ D = {b}^{2} - 4ac = ( - 3) {}^{2} - 4 \times 2 \times 1 = 9 - 8 = 1 \\ x_{1} = \frac{3 + 1}{2 \times 2} = \frac{4}{4} = 1 \\ x_{2} = \frac{3 - 1}{ 2\times 2} = \frac{2}{4} = 0.5 \\ 2 {x}^{2} - 3x + 1 = 2(x - 1)(x - 0.5)[/tex]

[tex]2 {x}^{2} - 3x + 1 > 0 \\ 2(x - 1)(x - 0.5) > 0 \\ + + + (0.5) - - - (1) + + + \\ otvet \: \: \: x \: \epsilon \: ( - \propto; \: 0.5)U(1; \: + \propto)[/tex]