document.write( "Question 108056This question is from textbook Introductoru & Intermediate Algebra
\n" ); document.write( ": The length of a rectangle exceeds twice its width by 3 inces. If the area is 10 square inches, find the rectangles dimensions. Round to the nearest tenth of an inch. \n" ); document.write( "
Algebra.Com's Answer #78715 by checkley71(8403)\"\" \"About 
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L=2W+3
\n" ); document.write( "L*W=10
\n" ); document.write( "(2W+3)*W=10
\n" ); document.write( "2W*2+3W-10=0
\n" ); document.write( "USING THE QUADRATIC EQUATON WE GET
\n" ); document.write( "\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"
\n" ); document.write( "X=(-3+-SQRT[3^2-4*2*-10])/2*2
\n" ); document.write( "X=(-3+-SQRT[9+80])/4
\n" ); document.write( "X=(-3+-SQRT89)/4
\n" ); document.write( "X=(-3+-9.43)/4
\n" ); document.write( "X=(-3+9.43)/4
\n" ); document.write( "X=6.43/4
\n" ); document.write( "X=1.6 ANSWER FOR THE WIDTH.
\n" ); document.write( "L=2*1.6
\n" ); document.write( "L=3.2+3
\n" ); document.write( "L=6.2 FOR THE LENGTH.
\n" ); document.write( "PROOF
\n" ); document.write( "1.6*6.2=10
\n" ); document.write( "10=10\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
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