document.write( "Question 847183: The area of a rectangle is 44 yards (squared) , and the length of the rectangle is 3 yd less than twice the width. Find the dimensions of the rectangle. \n" ); document.write( "
Algebra.Com's Answer #510248 by swincher4391(1107)\"\" \"About 
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A = L*W\r
\n" ); document.write( "\n" ); document.write( "L = 2W - 3\r
\n" ); document.write( "\n" ); document.write( "A = (2W-3)*W\r
\n" ); document.write( "\n" ); document.write( "44 = 2W^2 - 3W\r
\n" ); document.write( "\n" ); document.write( "2W^2 - 3W - 44 = 0\r
\n" ); document.write( "\n" ); document.write( "((3 +- sqrt((-3)^2 - 4*2*-44))/4 \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "(3 +- sqrt( 9 + 352))/4\r
\n" ); document.write( "\n" ); document.write( "(3 +- 19)/4\r
\n" ); document.write( "\n" ); document.write( "W= -16/4 and 22/4\r
\n" ); document.write( "\n" ); document.write( "However, W must be positive as we are talking about a measurement.\r
\n" ); document.write( "\n" ); document.write( "So W = 22/4 = 5.5\r
\n" ); document.write( "\n" ); document.write( "Plug that back into our original 44 = L * 5.5\r
\n" ); document.write( "\n" ); document.write( "44/5.5 = 8\r
\n" ); document.write( "\n" ); document.write( "So our dimensions are 8 x 5.5.
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