2x^2-x-3\le 0 x \in [-1;1\frac 12] x \in (-1;1\frac 12) x \in (-\infty; -1] \cup [1\frac 12;+\infty) x \in [1\frac 12; +\infty)

Задание

Реши неравенство

\(2x^2-x-3\le 0\)

\(x \in [-1;1\frac 12]\)
\(x \in (-1;1\frac 12)\)
\(x \in (-\infty; -1] \cup [1\frac 12;+\infty)\)
\( x \in [1\frac 12; +\infty)\)