document.write( "Question 214890: Please help me this word problem: Kevin drove 320 miles to a mountain resort. His return trip took 20 minutes longer because his speed returning was 4 mi/h slower than his speed going. \r
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Algebra.Com's Answer #162341 by checkley77(12844)\"\" \"About 
You can put this solution on YOUR website!
D=RT
\n" ); document.write( "320=RT OR R=320/T
\n" ); document.write( "320=(R-4)(T+1/3)
\n" ); document.write( "320=(320/T-4)(T+1/3)
\n" ); document.write( "320=320-4T+320/3T-4/3
\n" ); document.write( "-4T+320/3T-4/3=0
\n" ); document.write( "(-4T*3T+320-4T)/3T=0
\n" ); document.write( "(-12T^2+320-4T)/3T=0
\n" ); document.write( "12T^2+4T-320=0
\n" ); document.write( "4(3T^2+T-80)=0
\n" ); document.write( "\"T+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"
\n" ); document.write( "T=(-1+-SQRT[1^1-4*3*-80])/2*3
\n" ); document.write( "T=(-1+-SQRT[1+960])/6
\n" ); document.write( "T=(-1+-SQRT961)/6
\n" ); document.write( "T=(-1+-31)/6
\n" ); document.write( "T=-32/3=-5.333
\n" ); document.write( "T=(-1+31)/6
\n" ); document.write( "T=30/6
\n" ); document.write( "T=5 HRS. GOING.
\n" ); document.write( "PROOF;
\n" ); document.write( "320=R*5
\n" ); document.write( "R=320/5
\n" ); document.write( "R=64 MPH. GOING
\n" ); document.write( "320=(64-4)(5+1/3)
\n" ); document.write( "320=60*16/3
\n" ); document.write( "320=960/3
\n" ); document.write( "960=960
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