1. **State the problem:** Solve the quadratic equation $$-2x^2 + 12x - 5 = 0$$ by finding the discriminant, describing the roots, and finding the exact solutions using the quadratic formula. 2. **Recall the quadratic formula:** For an equation $$ax^2 + bx + c = 0$$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where the discriminant $$\Delta = b^2 - 4ac$$ determines the nature of the roots. 3. **Identify coefficients:** Here, $$a = -2$$, $$b = 12$$, and $$c = -5$$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 12^2 - 4 \times (-2) \times (-5) = 144 - 40 = 104$$ 5. **Interpret the discriminant:** Since $$\Delta = 104 > 0$$, there are two distinct real roots. 6. **Apply the quadratic formula:** $$x = \frac{-12 \pm \sqrt{104}}{2 \times (-2)} = \frac{-12 \pm \sqrt{104}}{-4}$$ 7. **Simplify the square root:** $$\sqrt{104} = \sqrt{4 \times 26} = 2\sqrt{26}$$ 8. **Substitute back:** $$x = \frac{-12 \pm 2\sqrt{26}}{-4}$$ 9. **Simplify the fraction by canceling common factors:** $$x = \frac{\cancel{2}(-6 \pm \sqrt{26})}{\cancel{2}(-2)} = \frac{-6 \pm \sqrt{26}}{-2}$$ 10. **Divide numerator and denominator by -1 to simplify sign:** $$x = \frac{6 \mp \sqrt{26}}{2}$$ 11. **Final exact solutions:** $$x_1 = \frac{6 + \sqrt{26}}{2}, \quad x_2 = \frac{6 - \sqrt{26}}{2}$$