```Question 180501
A business invests \$8,000 in a savings account for two years.  At the beginning of the second year, an additional \$2,500 is invested.  At the end of the second year, the account balance is \$11,445.  What was the annual interest rate?
:
Let x = interest rate in decimal form
:
8000(x+1) = 1st year total
or
8000x + 8000
:
2nd year total
(x+1)(8000 + 8000x + 2500) = 11445
(x+1)(8000x + 10500) = 11445
FOIL
8000x^2 + 10500x + 8000x + 10500 - 11445 = 0
:
8000x^2 + 18500x - 945 = 0
:
simplify divide equation by 5
1600x^2 + 3700x - 189 = 0
:
Use the quadratic formula to solve this:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
:
a=1600; b=3700, c=-189
{{{x = (-3700 +- sqrt(3700^2 - 4 * 1600 * -189 ))/(2*1600) }}}
{{{x = (-3700 +- sqrt(13690000 + 1209600 ))/(3200) }}}
{{{x = (-3700 +- sqrt(14899600 ))/(3200) }}}
Positive solution is what we want here:
{{{x = (-3700 + 3860)/(3200) }}}
{{{x = 160/3200}}}
x = .05,  Therefore 5% is the interest rate
;
:
Check solution:
1st year 1.05*8000 = 8400
2nd year 1.05(8400 + 2500) = 11445, confirms our solution
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