1. **State the problem:** Find the x-intercepts of the parabola given by the equation $$y = -2x^2 + 6x - 4$$. The x-intercepts occur where $$y=0$$. 2. **Set the equation to zero:** $$0 = -2x^2 + 6x - 4$$ 3. **Use the quadratic formula:** For a quadratic equation $$ax^2 + bx + c = 0$$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a = -2$$, $$b = 6$$, and $$c = -4$$. 4. **Calculate the discriminant:** $$b^2 - 4ac = 6^2 - 4(-2)(-4) = 36 - 32 = 4$$ 5. **Calculate the roots:** $$x = \frac{-6 \pm \sqrt{4}}{2(-2)} = \frac{-6 \pm 2}{-4}$$ 6. **Find each root separately:** - For the plus sign: $$x_1 = \frac{-6 + 2}{-4} = \frac{\cancel{-6} + 2}{\cancel{-4}} = \frac{-4}{-4} = 1$$ - For the minus sign: $$x_2 = \frac{-6 - 2}{-4} = \frac{-8}{-4} = 2$$ 7. **Write the x-intercepts as ordered pairs:** Since $$y=0$$ at the x-intercepts, the points are: $$(1, 0), (2, 0)$$ **Final answer:** The x-intercepts of the parabola are (1, 0) and (2, 0).