Question:
A worker receives an annual wage of $\$20{,}000$, which he always deposits into a savings account at the end of the year. By the end of the third year (when he makes the third deposit), he wants to have at least $\$66,200$ in the account to finance the purchase of a house. What is the minimal compound interest rate that the savings account must provide? Express your answer as a percentage, but do not include the percent sign.

Answer:
If the interest rate is $r$, it follows that $$20000(1+r)^2 + 20000(1+r) + 20000 \ge 66200.$$ If we set $x = 1+r$ and divide through the inequality by $200$, it follows that $$100x^2 + 100x - 231 \ge 0.$$ Since $231 = 11 \cdot 21$, we can factor the quadratic as $(10x - 11)(10x + 21) \ge 0$, so it follows that $x \ge \frac {11}{10}$ or $x \le \frac{-21}{10}$. Since we are looking for an interest rate percentage, it follows that $x \ge \frac{11}{10} = 1.1$, and $r = x - 1 = \boxed{10}\%$.