1. **State the problem:** Find the solutions to the system of equations where a parabola and a line intersect. 2. **Given equations:** Assume the parabola is $y = -a(x - 4)^2 + b$ and the line is $y = mx + c$. 3. **Set the equations equal to find intersection points:** $$-a(x - 4)^2 + b = mx + c$$ 4. **Rewrite and expand:** $$-a(x^2 - 8x + 16) + b = mx + c$$ $$-a x^2 + 8a x - 16a + b = mx + c$$ 5. **Bring all terms to one side:** $$-a x^2 + 8a x - 16a + b - mx - c = 0$$ 6. **Group like terms:** $$-a x^2 + (8a - m) x + (b - 16a - c) = 0$$ 7. **Use quadratic formula:** $$x = \frac{-(8a - m) \pm \sqrt{(8a - m)^2 - 4(-a)(b - 16a - c)}}{2(-a)}$$ 8. **Simplify denominator:** $$x = \frac{-(8a - m) \pm \sqrt{(8a - m)^2 + 4a(b - 16a - c)}}{-2a}$$ 9. **Calculate discriminant:** $$\Delta = (8a - m)^2 + 4a(b - 16a - c)$$ 10. **Interpret solutions:** - If $\Delta > 0$, two real solutions. - If $\Delta = 0$, one real solution. - If $\Delta < 0$, no real solutions. **Final answer:** The solutions for $x$ are given by the quadratic formula above, and corresponding $y$ values are found by substituting back into $y = mx + c$ or $y = -a(x - 4)^2 + b$.