1. **State the problem:** Solve the quadratic equation $9n^2 + 79 = -18n$ by completing the square. 2. **Rewrite the equation:** Move all terms to one side to set the equation to zero: $$9n^2 + 18n + 79 = 0$$ 3. **Divide through by the coefficient of $n^2$ to simplify:** $$\cancel{9}n^2 + \cancel{9} \cdot 2n + \frac{79}{9} = 0 \implies n^2 + 2n + \frac{79}{9} = 0$$ 4. **Isolate the constant term:** $$n^2 + 2n = -\frac{79}{9}$$ 5. **Complete the square:** Take half of the coefficient of $n$, which is $2$, half is $1$, then square it: $1^2 = 1$. Add $1$ to both sides: $$n^2 + 2n + 1 = -\frac{79}{9} + 1$$ 6. **Simplify the right side:** $$-\frac{79}{9} + \frac{9}{9} = -\frac{70}{9}$$ 7. **Rewrite the left side as a perfect square:** $$(n + 1)^2 = -\frac{70}{9}$$ 8. **Solve for $n$ by taking the square root of both sides:** $$n + 1 = \pm \sqrt{-\frac{70}{9}} = \pm \frac{\sqrt{-70}}{3} = \pm \frac{\sqrt{70}i}{3}$$ 9. **Isolate $n$:** $$n = -1 \pm \frac{\sqrt{70}i}{3}$$ **Final answer:** $$n = -1 \pm \frac{\sqrt{70}i}{3}$$ This means the solutions are complex numbers because the discriminant is negative.