1. **State the problem:** Solve the quadratic equation $$-6x^{2} + 17x - 5 = 0$$. 2. **Formula used:** The quadratic formula for solving $$ax^{2} + bx + c = 0$$ is $$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$. 3. **Identify coefficients:** Here, $$a = -6$$, $$b = 17$$, and $$c = -5$$. 4. **Calculate the discriminant:** $$\Delta = b^{2} - 4ac = 17^{2} - 4(-6)(-5) = 289 - 120 = 169$$. 5. **Evaluate the square root of the discriminant:** $$\sqrt{169} = 13$$. 6. **Apply the quadratic formula:** $$x = \frac{-17 \pm 13}{2(-6)} = \frac{-17 \pm 13}{-12}$$. 7. **Calculate the two solutions:** - For the plus sign: $$x = \frac{-17 + 13}{-12} = \frac{-4}{-12} = \frac{1}{3}$$. - For the minus sign: $$x = \frac{-17 - 13}{-12} = \frac{-30}{-12} = \frac{5}{2}$$. **Final answer:** The solutions to the equation are $$x = \frac{1}{3}$$ and $$x = \frac{5}{2}$$.