document.write( "Question 128995: I don't understand this problem can someone please help me. The area of a rectangle is 45 square cm. If the length is 4 cm greater than the width, what are the dimensions of the rectangle?.
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Algebra.Com's Answer #94329 by oscargut(2103)\"\" \"About 
You can put this solution on YOUR website!
you have to find the length and the width of the rectangle
\n" ); document.write( "let denotate L=length and W=width\r
\n" ); document.write( "\n" ); document.write( "L=W+4\r
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\n" ); document.write( "\n" ); document.write( "area = L*W =45 then (W+4)W=45 then W^2+4W-45=0
\n" ); document.write( "then \"W+=+%28-4+%2B-+sqrt%28+%284%29%5E2-4%2A1%2A-45+%29%29%2F%282%2A1%29+\"
\n" ); document.write( "= \"+%28-4+%2B-+sqrt%28196%29%29%2F%282%2A1%29+\"
\n" ); document.write( "=\"+%28-4+%2B-+14%29%2F2+\"
\n" ); document.write( "=\"+%28-2+%2B-+7%29+\"
\n" ); document.write( "solutions are W=-9 and W=5 then W=5\r
\n" ); document.write( "\n" ); document.write( "then W=5 and L=9 \r
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