1. **State the problem:** Solve the equation $$\frac{2}{x} = 2x + 1$$ for $x$. 2. **Rewrite the equation:** To eliminate the fraction, multiply both sides by $x$ (assuming $x \neq 0$): $$2 = x(2x + 1)$$ 3. **Expand the right side:** $$2 = 2x^2 + x$$ 4. **Bring all terms to one side to form a quadratic equation:** $$2x^2 + x - 2 = 0$$ 5. **Use the quadratic formula:** For $ax^2 + bx + c = 0$, solutions are $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=2$, $b=1$, $c=-2$. 6. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 1^2 - 4 \times 2 \times (-2) = 1 + 16 = 17$$ 7. **Find the roots:** $$x = \frac{-1 \pm \sqrt{17}}{2 \times 2} = \frac{-1 \pm \sqrt{17}}{4}$$ 8. **Final answer:** $$x = \frac{-1 + \sqrt{17}}{4} \quad \text{or} \quad x = \frac{-1 - \sqrt{17}}{4}$$