1. **State the problem:** Solve the inequality $$2x^2 + x - 1 > 0$$. 2. **Identify the type of inequality:** This is a quadratic inequality. To solve it, we first find the roots of the quadratic equation $$2x^2 + x - 1 = 0$$. 3. **Use the quadratic formula:** For $$ax^2 + bx + c = 0$$, the roots are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Here, $$a=2$$, $$b=1$$, and $$c=-1$$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 1^2 - 4 \times 2 \times (-1) = 1 + 8 = 9$$. 5. **Find the roots:** $$x = \frac{-1 \pm \sqrt{9}}{2 \times 2} = \frac{-1 \pm 3}{4}$$ So, $$x_1 = \frac{-1 - 3}{4} = \frac{-4}{4} = -1$$ $$x_2 = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}$$ 6. **Analyze the sign of the quadratic:** Since $$a=2 > 0$$, the parabola opens upwards. 7. **Determine intervals where $$2x^2 + x - 1 > 0$$:** The quadratic is positive outside the roots and negative between them. Therefore, the solution is: $$x < -1 \quad \text{or} \quad x > \frac{1}{2}$$. **Final answer:** $$\boxed{x < -1 \text{ or } x > \frac{1}{2}}$$.