SOLUTION: Could anyone help me explain set-up of this equation: A business invests $10,000 in a savings account for two years. At the beginning of the second year, an additional $3500 is

Question 149764: Could anyone help me explain set-up of this equation:
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?
Answer by vleith(2983) (Show Source): You can put this solution on YOUR website!
Let x be the interest rate. For example if x = 9% , then use x =0.09
You invest 10000 at rate x. After one year you will have
You then add 3500 and let the sum accumulate interest for another year.

At the end of the year you are told you have 15569.75
So

use the quadratic equation to solve for x

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=635040000 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.085, -2.435. Here's your graph:

The end amount is greater than the amount you put in, so the interest rate must be positive.
The positive answer is 0.085. Which results in a return of 8.5%.
Check your answer by plugging in 0.085 and verifying the result

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