An investor made an investment and deposited $\$ 2,000$ in an account at a certain annual interest rate. One year later, he deposited $\$ 3,000$ in an investment in another account at the same rate. At the end of the second year, he accumulated a total value of $\$ 5,247.45$ from these two investments. What is the interest rate? Note: Input your answer as an integer with one decimal value with \% sign (no space). Round to nearest tenth, for example: $2.7 \%$.

Step 1 ：Let the interest rate be r. Then, the first investment grows to $2000(1+r)$ after one year, and the second investment grows to $3000(1+r)$ after one year.

Step 2 ：At the end of the second year, the first investment grows to $2000(1+r)^2$, and the second investment grows to $3000(1+r)$.

Step 3 ：The total value of the investments at the end of the second year is $2000(1+r)^2 + 3000(1+r) = 5247.45$.

Step 4 ：Let x = 1+r. Then, the equation becomes $2000x^2 + 3000x = 5247.45$.

Step 5 ：Divide the equation by $1000$ to simplify: $2x^2 + 3x = 5.24745$.

Step 6 ：Subtract $5.24745$ from both sides: $2x^2 + 3x - 5.24745 = 0$.

Step 7 ：Use the quadratic formula to solve for x: $x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-5.24745)}}{4}$.

Step 8 ：Calculate the two possible values for x: $x \approx 1.025$ or $x \approx -2.475$. Since x represents the interest rate, we can ignore the negative value.

Step 9 ：Now, we have $x \approx 1.025$. To find the interest rate r, subtract 1 from x: $r \approx 1.025 - 1 = 0.025$.

Step 10 ：Convert r to a percentage and round to the nearest tenth: $r \approx 2.5\%$.

Step 11 ：\(\boxed{2.5\%}\)