Solve the nonlinear inequality.Express the solution using interv 2x^2+xgeqslant 6 square

To solve the nonlinear inequality \(2x^2 + x \geq 6\), we first rewrite it in standard form:<br /><br />\[2x^2 + x - 6 \geq 0\]<br /><br />Next, we solve the corresponding quadratic equation \(2x^2 + x - 6 = 0\) to find the critical points. We can use the quadratic formula:<br /><br />\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]<br /><br />For the quadratic equation \(2x^2 + x - 6 = 0\), \(a = 2\), \(b = 1\), and \(c = -6\). Plugging these values into the quadratic formula gives:<br /><br />\[x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-6)}}{2(2)}\]<br />\[x = \frac{-1 \pm \sqrt{1 + 48}}{4}\]<br />\[x = \frac{-1 \pm \sqrt{49}}{4}\]<br />\[x = \frac{-1 \pm 7}{4}\]<br /><br />This gives us two solutions:<br /><br />\[x = \frac{-1 + 7}{4} = \frac{6}{4} = \frac{3}{2}\]<br />\[x = \frac{-1 - 7}{4} = \frac{-8}{4} = -2\]<br /><br />Now, we test the intervals determined by these critical points: \((-\infty, -2)\), \((-2, \frac{3}{2})\), and \((\frac{3}{2}, \infty)\).<br /><br />For \(x \in (-\infty, -2)\), choose \(x = -3\):<br />\[2(-3)^2 + (-3) - 6 = 18 - 3 - 6 = 9 > 0\]<br /><br />For \(x \in (-2, \frac{3}{2})\), choose \(x = 0\):<br />\[2(0)^2 + (0) - 6 = -6 < 0\]<br /><br />For \(x \in (\frac{3}{2}, \infty)\), choose \(x = 2\):<br />\[2(2)^2 + (2) - 6 = 8 + 2 - 6 = 4 > 0\]<br /><br />Therefore, the solution to the inequality \(2x^2 + x \geq 6\) is:<br /><br />\[\boxed{(-\infty, -2) \cup \left(\frac{3}{2}, \infty\right)}\]