document.write( "Question 89562: The length of a rectangle is 2 cm more than 5 times its width. If the area of the rectangle is 65 cm2, find the dimensions of the rectangle to the nearest thousandth\r
\n" ); document.write( "\n" ); document.write( "Hint: Call the width x. Then the length is 5x + 2. Now write your equation and solve.
\n" ); document.write( "

Algebra.Com's Answer #65101 by checkley75(3666)\"\" \"About 
You can put this solution on YOUR website!
L=5W+2 & THE AREA=W(5W+2)
\n" ); document.write( "5W^2+2W=65
\n" ); document.write( "5W^2+2W-65=0
\n" ); document.write( "USING THE QUADRATIC EQUATION WE GET: \"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"
\n" ); document.write( "X=(-2+-SQRT[2^2-4*5*-65])/2*5
\n" ); document.write( "X=(-2+-SQRT[4+1300])/10
\n" ); document.write( "X=(-2+-SQRT1304)/10
\n" ); document.write( "X=(-2+-36.11)/10
\n" ); document.write( "X=(-2+36.11)/10
\n" ); document.write( "X=34.11)/10
\n" ); document.write( "X=3.41 ANSWER FOR THE WIDTH.
\n" ); document.write( "L=5*3.41+2
\n" ); document.write( "L=17.05+2
\n" ); document.write( "L=19.05 ANSWER FOR THE LENGTH.
\n" ); document.write( "PROOF
\n" ); document.write( "19.05*3.41=65
\n" ); document.write( "65=65
\n" ); document.write( "
\n" );