1. **State the problem:** Solve the system of equations by graphing: $$h = d^2 - 16d + 60$$ $$h = 12d - 55$$ 2. **Understand the system:** The first equation is a quadratic function (a parabola), and the second is a linear function (a straight line). 3. **Set the equations equal to find intersection points:** Since both equal $h$, set them equal: $$d^2 - 16d + 60 = 12d - 55$$ 4. **Bring all terms to one side:** $$d^2 - 16d + 60 - 12d + 55 = 0$$ $$d^2 - 28d + 115 = 0$$ 5. **Solve the quadratic equation:** Use the quadratic formula: $$d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-28$, $c=115$. Calculate the discriminant: $$\Delta = (-28)^2 - 4 \times 1 \times 115 = 784 - 460 = 324$$ 6. **Find the roots:** $$d = \frac{28 \pm \sqrt{324}}{2} = \frac{28 \pm 18}{2}$$ So, $$d_1 = \frac{28 + 18}{2} = \frac{46}{2} = 23$$ $$d_2 = \frac{28 - 18}{2} = \frac{10}{2} = 5$$ 7. **Find corresponding $h$ values:** Substitute $d$ into the linear equation $h = 12d - 55$: For $d=23$: $$h = 12 \times 23 - 55 = 276 - 55 = 221$$ For $d=5$: $$h = 12 \times 5 - 55 = 60 - 55 = 5$$ 8. **Final solution:** The system has two solutions: $$(d,h) = (23, 221) \text{ and } (5, 5)$$ These are the points where the parabola and the line intersect.