document.write( "Question 193776: 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? \n" ); document.write( "
Algebra.Com's Answer #145460 by Mathtut(3670)\"\" \"About 
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the annual interest rate = r
\n" ); document.write( "$8,000 in a savings account for two years.--> 8000*(1+r)^2
\n" ); document.write( "beginning of the second year, an additional $2,500 is invested-->2500*(100+r)
\n" ); document.write( "8000*(1+r)^2 + 2500*(1+r) = 11,445
\n" ); document.write( "let 100+r = A
\n" ); document.write( ":\r
\n" ); document.write( "\n" ); document.write( "8000*A^2 + 2500*A = 11,445:\r
\n" ); document.write( "\n" ); document.write( "8000*A^2 + 2500*A - 11,445 = 0:\r
\n" ); document.write( "\n" ); document.write( "A = 1.05
\n" ); document.write( ":
\n" ); document.write( "r = 0.05 or 5 % Ans.
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