document.write( "Question 113765This question is from textbook Beginning Algebra
\n" ); document.write( ": The length of a rectangle is 3cm more than 2 times the width. If the area of the rectangle is 99 cm^2, find the dimensions of the rectangle to the nearest thousandth. \r
\n" ); document.write( "\n" ); document.write( "The equation I got from this is:
\n" ); document.write( "x(2x+3)=99\r
\n" ); document.write( "\n" ); document.write( "I have worked it to this point, and gotten stuck.
\n" ); document.write( "2x^2+3x=99
\n" ); document.write( "2x^2+3x-99=0\r
\n" ); document.write( "\n" ); document.write( "I am not sure what to do after this. I have tried to factor this, but am not able to. What do I do now?
\n" ); document.write( "

Algebra.Com's Answer #82768 by checkley71(8403)\"\" \"About 
You can put this solution on YOUR website!
L=3+2W
\n" ); document.write( "AREA=L*W
\n" ); document.write( "99=(3+2W)W
\n" ); document.write( "99=3W+2W^2
\n" ); document.write( "2W^2+3W-99=0
\n" ); document.write( "USING THE QUADRATIC EQUATION \"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"
\n" ); document.write( "YOU GET:
\n" ); document.write( "W=(-3+-SQRT[3^3-4*2*-99])/2*2
\n" ); document.write( "W=(-3+-SQRT[9+792])/4
\n" ); document.write( "W=(-3+-SQRT801)/4
\n" ); document.write( "W=(-3+-28.3)/4
\n" ); document.write( "W=(-3+28.3)/4
\n" ); document.write( "W=25.3/4
\n" ); document.write( "W=6.325 FOR THE WIDTH.
\n" ); document.write( "L=3+2*6.325
\n" ); document.write( "L=3+12.65
\n" ); document.write( "L=15.65 FOR THE LENGTH.
\n" ); document.write( "PROOF:
\n" ); document.write( "99=(3+2*6.325)*6.325
\n" ); document.write( "99=(3+12.65)*6.325
\n" ); document.write( "99=15.65*6.325
\n" ); document.write( "99~99\r
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