1. **State the problem:** Solve the system of equations: $$y = x^2 - 4x - 4$$ $$y = -2x^2 + 4x - 1$$ 2. **Set the two expressions for y equal to each other** since both equal y: $$x^2 - 4x - 4 = -2x^2 + 4x - 1$$ 3. **Bring all terms to one side to form a quadratic equation:** $$x^2 - 4x - 4 + 2x^2 - 4x + 1 = 0$$ 4. **Combine like terms:** $$3x^2 - 8x - 3 = 0$$ 5. **Use the quadratic formula to solve for x:** The quadratic formula is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=3$, $b=-8$, and $c=-3$. 6. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (-8)^2 - 4 \times 3 \times (-3) = 64 + 36 = 100$$ 7. **Calculate the roots:** $$x = \frac{-(-8) \pm \sqrt{100}}{2 \times 3} = \frac{8 \pm 10}{6}$$ 8. **Find the two solutions for x:** - For $x = \frac{8 + 10}{6} = \frac{18}{6} = 3$ - For $x = \frac{8 - 10}{6} = \frac{-2}{6} = -\frac{1}{3}$ 9. **Find corresponding y values by substituting back into one of the original equations, e.g.,** $$y = x^2 - 4x - 4$$ - For $x=3$: $$y = 3^2 - 4 \times 3 - 4 = 9 - 12 - 4 = -7$$ - For $x = -\frac{1}{3}$: $$y = \left(-\frac{1}{3}\right)^2 - 4 \times \left(-\frac{1}{3}\right) - 4 = \frac{1}{9} + \frac{4}{3} - 4 = \frac{1}{9} + \frac{12}{9} - \frac{36}{9} = -\frac{23}{9}$$ 10. **Final solution:** $$\boxed{(3, -7) \text{ and } \left(-\frac{1}{3}, -\frac{23}{9}\right)}$$