1. **State the problem:** Solve the quadratic expression $w^2 - 15w + 54 = 0$ for $w$. 2. **Formula and rules:** To solve a quadratic equation $ax^2 + bx + c = 0$, we can factor it if possible or use the quadratic formula: $$w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=1$, $b=-15$, and $c=54$. 3. **Try factoring:** We look for two numbers that multiply to $54$ and add to $-15$. These numbers are $-9$ and $-6$ because $-9 \times -6 = 54$ and $-9 + (-6) = -15$. 4. **Factor the quadratic:** $$w^2 - 15w + 54 = (w - 9)(w - 6) = 0$$ 5. **Solve each factor:** Set each factor equal to zero: $$w - 9 = 0 \Rightarrow w = 9$$ $$w - 6 = 0 \Rightarrow w = 6$$ 6. **Final answer:** The solutions to the quadratic equation are $$w = 9 \text{ or } w = 6$$