1. The problem asks us to find all values of $k$ such that the quadratic equation $3x^2 - 18x + k = 0$ has real, unequal roots. 2. Recall that for a quadratic equation $ax^2 + bx + c = 0$, the discriminant $\Delta$ is given by: $$\Delta = b^2 - 4ac$$ 3. The roots are real and unequal if and only if the discriminant is positive: $$\Delta > 0$$ 4. For the given equation, $a = 3$, $b = -18$, and $c = k$. Substitute these into the discriminant formula: $$\Delta = (-18)^2 - 4 \times 3 \times k = 324 - 12k$$ 5. Set the discriminant greater than zero to find the range of $k$: $$324 - 12k > 0$$ 6. Solve the inequality: $$324 > 12k$$ $$\frac{324}{12} > k$$ $$27 > k$$ 7. Therefore, the quadratic equation has real, unequal roots for all values of $k$ such that: $$k < 27$$ Final answer: $k < 27$