# Thread: Find Annual Interest Rate

1. ## Find Annual Interest Rate

A business invests $10,000 in a savings account for two years. At the beginning of the second year, an additional$3500 is invested. At the end of the second year, the account balance is $15,569.75. What was the annual interest rate? Is there a formula? 2.$\displaystyle
\begin{gathered}
10,000\left( {1 + i} \right)^2 + 3,500\left( {1 + i} \right) = 15,569.75 \hfill \\
\Leftrightarrow \hfill \\
\end{gathered}
\displaystyle
\begin{gathered}
\left[ {10,000\left( {1 + i} \right) + 3,500} \right]\left( {1 + i} \right) = 15,569.75 \hfill \\
\Leftrightarrow \hfill \\
\end{gathered}
\displaystyle
\begin{gathered}
\left[ {10,000 + 10,000i + 3,500} \right]\left( {1 + i} \right) = 15,569.75 \hfill \\
\Leftrightarrow \hfill \\
\end{gathered}
\displaystyle
10,000 + 10,000i + 3,500 + 10,000i + 10,000i^2 + 3,500i = 15,569.75
$You need the quadratic formula. 3. ## ok.... Originally Posted by jonah$\displaystyle
\begin{gathered}
10,000\left( {1 + i} \right)^2 + 3,500\left( {1 + i} \right) = 15,569.75 \hfill \\
\Leftrightarrow \hfill \\
\end{gathered}
\displaystyle
\begin{gathered}
\left[ {10,000\left( {1 + i} \right) + 3,500} \right]\left( {1 + i} \right) = 15,569.75 \hfill \\
\Leftrightarrow \hfill \\
\end{gathered}
\displaystyle
\begin{gathered}
\left[ {10,000 + 10,000i + 3,500} \right]\left( {1 + i} \right) = 15,569.75 \hfill \\
\Leftrightarrow \hfill \\
\end{gathered}
\displaystyle
10,000 + 10,000i + 3,500 + 10,000i + 10,000i^2 + 3,500i = 15,569.75
\$