document.write( "Question 1128472: Use the compound interest formula Upper A= P( 1 + r )^t to find the annual interest​ rate, r, if in 2 years an investment of $2,000 grows to $ 3,380.\r
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Algebra.Com's Answer #744927 by MathLover1(20850)\"\" \"About 
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\n" ); document.write( "\"+A=+P%28+1+%2B+r+%29%5Et\" \r
\n" ); document.write( "\n" ); document.write( "\"t=2\" years \r
\n" ); document.write( "\n" ); document.write( "\"P=2000\" grows to \r
\n" ); document.write( "\n" ); document.write( "\"A=3380\"\r
\n" ); document.write( "\n" ); document.write( "\"r\"=?\r
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\n" ); document.write( "\n" ); document.write( "\"+3380=+2000%28+1+%2B+r+%29%5E2\" \r
\n" ); document.write( "\n" ); document.write( "\"+3380%2F2000=%28+1+%2B+r+%29%5E2\" \r
\n" ); document.write( "\n" ); document.write( "\"+1.69=1+%2B+2r+%2Br%5E2\" \r
\n" ); document.write( "\n" ); document.write( "\"+r%5E2%2B+2r+%2B1-1.69=0\" \r
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\n" ); document.write( "\n" ); document.write( "\"+r%5E2%2B+2r-0.69=0\" \r
\n" ); document.write( "\n" ); document.write( "\"+100r%5E2%2B+200r-69=0\" \r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28-200+%2B-+sqrt%28+200%5E2-4%2A100%2A%28-69%29+%29%29%2F%282%2A100%29+\" \r
\n" ); document.write( "\n" ); document.write( "\"r+=+%28-200+%2B-+sqrt%28+40000%2B27600+%29%29%2F200+\" \r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28-200+%2B-+sqrt%28+67600+%29%29%2F200+\" \r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28-200+%2B-+260%29%2F200+\" \r
\n" ); document.write( "\n" ); document.write( "\"r+=+%28-10+%2B-+13%29%2F10+\" ...need only positive solution\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28-10+%2B+13%29%2F10+\" \r
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\n" ); document.write( "\n" ); document.write( "\"r+=+3%2F10\"\r
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\n" ); document.write( "\n" ); document.write( "\"r=0.3\" => \"r=30\"%\r
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