document.write( "Question 845646: The area of a rectangle is 40. If the length is 6 more than the width, what are the dimensions of the rectangle.
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Algebra.Com's Answer #509395 by jim_thompson5910(35256)\"\" \"About 
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x(x+6) = 40\r
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\n" ); document.write( "\n" ); document.write( "x^2 + 6x = 40\r
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\n" ); document.write( "\n" ); document.write( "x^2 + 6x - 40 = 0\r
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\n" ); document.write( "\n" ); document.write( "(x + 10)(x - 4) = 0\r
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\n" ); document.write( "\n" ); document.write( "x + 10 = 0 or x - 4 = 0\r
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\n" ); document.write( "\n" ); document.write( "x = -10 or x = 4\r
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\n" ); document.write( "\n" ); document.write( "Toss the negative result.\r
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\n" ); document.write( "\n" ); document.write( "The only solution is x = 4\r
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\n" ); document.write( "\n" ); document.write( "x = 4\r
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\n" ); document.write( "\n" ); document.write( "x + 6 = 4+6 = 10\r
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\n" ); document.write( "\n" ); document.write( "The dimensions are\r
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\n" ); document.write( "\n" ); document.write( "Length: 10
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