Answer
$$\frac{-\sqrt{17}-7}{4}\le \:x\le \frac{\sqrt{17}-7}{4}$$
Work Step by Step
Solving the inequality using the rules of inequalities, we find:
$$-2x^2-7x-4\ge \:0 \\ -2\left(x+\frac{7}{4}\right)^2+\frac{17}{8}\ge \:0 \\ \left(x+\frac{7}{4}\right)^2\le \frac{17}{16} \\ \frac{-\sqrt{17}-7}{4}\le \:x\le \frac{\sqrt{17}-7}{4}$$
