document.write( "Question 883362: The perimeter of a rectangle is 40 inches, and the area is 84 square inches. Find the length and width of the rectangle. \n" ); document.write( "
Algebra.Com's Answer #533566 by josgarithmetic(39618)\"\" \"About 
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x and y dimensions.
\n" ); document.write( "Perimeter,2x+2y=40 simplifiable to x+y=20.
\n" ); document.write( "Area, xy=84.\r
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\n" ); document.write( "\n" ); document.write( "Use substitution.
\n" ); document.write( "\"y=20-x\".
\n" ); document.write( "\"x%2820-x%29=84\".
\n" ); document.write( "\"-x%5E2%2B20x-84=0\"
\n" ); document.write( "\"highlight_green%28x%5E2-20x%2B84=0%29\".\r
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\n" ); document.write( "\n" ); document.write( "Discriminant: \"%28-20%29%5E2-4%2A84=64=8%5E2\"
\n" ); document.write( "Continue in the general solution for a quadratic equation:
\n" ); document.write( "\"x=%2820%2B8%29%2F2=14\", guessing that the PLUS form should work and not the MINUS form.\r
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\n" ); document.write( "\n" ); document.write( "\"y=20-14=6\"; but actually, either PLUS or MINUS form would work. One way would give 14 as found, and the other would give 6.\r
\n" ); document.write( "\n" ); document.write( "The dimensions are 14 and 6
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