Answer:
[tex]\{-\frac{1}{2} \} \cup \{\frac{2}{5}\}[/tex]
Step-by-step explanation:
[tex]10x^2 + x - 2 = 0; \\ a = 10, b = 1, c = -2; \\ D = b^2 - 4ac = 1^2 - 4 * 10 * (-2) = 1 + 80 = 81 = 9^2, > 0; \\ x_{1, 2} = \frac{-b \pm \sqrt{D}}{2a} = \frac{-1 \pm \sqrt{9^2}}{2 * 10} = \frac{-1 \pm 9}{20} = \left \ [ {{\frac{-1 - 9}{20} = -\frac{10}{20} = -\frac{1}{2} } \atop {\frac{-1 + 9}{20} = \frac{8}{20} = \frac{2}{5}}} \right.[/tex]
