Answer
$$\frac{-\sqrt{19}+3}{5}\le \:x\le \frac{\sqrt{19}+3}{5}$$
Work Step by Step
Solving the inequality using the rules of inequalities, we find:
$$\left(x-\frac{3}{5}\right)^2\le \frac{19}{25} \\ -\sqrt{\frac{19}{25}}\le \:x-\frac{3}{5}\le \sqrt{\frac{19}{25}} \\ \frac{-\sqrt{19}+3}{5}\le \:x\le \frac{\sqrt{19}+3}{5}$$
