# 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. $
\begin{gathered}
10,000\left( {1 + i} \right)^2 + 3,500\left( {1 + i} \right) = 15,569.75 \hfill \\
\Leftrightarrow \hfill \\
\end{gathered}
$

$
\begin{gathered}
\left[ {10,000\left( {1 + i} \right) + 3,500} \right]\left( {1 + i} \right) = 15,569.75 \hfill \\
\Leftrightarrow \hfill \\
\end{gathered}
$

$
\begin{gathered}
\left[ {10,000 + 10,000i + 3,500} \right]\left( {1 + i} \right) = 15,569.75 \hfill \\
\Leftrightarrow \hfill \\
\end{gathered}
$

$
10,000 + 10,000i + 3,500 + 10,000i + 10,000i^2 + 3,500i = 15,569.75
$

3. ## ok....

Originally Posted by jonah
$
\begin{gathered}
10,000\left( {1 + i} \right)^2 + 3,500\left( {1 + i} \right) = 15,569.75 \hfill \\
\Leftrightarrow \hfill \\
\end{gathered}
$

$
\begin{gathered}
\left[ {10,000\left( {1 + i} \right) + 3,500} \right]\left( {1 + i} \right) = 15,569.75 \hfill \\
\Leftrightarrow \hfill \\
\end{gathered}
$

$
\begin{gathered}
\left[ {10,000 + 10,000i + 3,500} \right]\left( {1 + i} \right) = 15,569.75 \hfill \\
\Leftrightarrow \hfill \\
\end{gathered}
$

$
10,000 + 10,000i + 3,500 + 10,000i + 10,000i^2 + 3,500i = 15,569.75
$