1. **Stating the problem:** Solve the quadratic equation $$-x^2 + 6x - 4 = 0$$ and find the turning point of the function $$f(x) = -x^2 + 6x - 4$$. 2. **Solving the quadratic equation:** The general form is $$ax^2 + bx + c = 0$$. Here, $$a = -1$$, $$b = 6$$, $$c = -4$$. Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 6^2 - 4(-1)(-4) = 36 - 16 = 20$$ 4. **Calculate the roots:** $$x = \frac{-6 \pm \sqrt{20}}{2(-1)} = \frac{-6 \pm 2\sqrt{5}}{-2}$$ Simplify numerator and denominator: $$x = \frac{\cancel{-6} \pm 2\sqrt{5}}{\cancel{-2}} = \frac{-6 \pm 2\sqrt{5}}{-2}$$ Dividing numerator and denominator by 2: $$x = \frac{\cancel{-6} \pm 2\sqrt{5}}{\cancel{-2}} = \frac{-3 \pm \sqrt{5}}{-1}$$ Multiply numerator and denominator by -1: $$x = 3 \mp \sqrt{5}$$ 5. **Calculate numerical values:** $$\sqrt{5} \approx 2.236$$ So, $$x_1 = 3 - 2.236 = 0.764 \approx 0.8$$ $$x_2 = 3 + 2.236 = 5.236 \approx 5.2$$ 6. **Finding the turning point:** For a quadratic $$f(x) = ax^2 + bx + c$$, the turning point $$x$$-coordinate is: $$x = -\frac{b}{2a} = -\frac{6}{2(-1)} = -\frac{6}{-2} = 3$$ Calculate $$f(3)$$: $$f(3) = -(3)^2 + 6(3) - 4 = -9 + 18 - 4 = 5$$ So the turning point is at $$(3, 5)$$. **Final answers:** - Solutions to the equation: $$x \approx 0.8$$ and $$x \approx 5.2$$. - Turning point coordinates: $$(3, 5)$$.