document.write( "Question 332220: The braking distance (in feet) of a car going V mph is given by \"d%28v%29=v%5E2%2F20%2Bv\" v is greater or equal to 0.\r
\n" ); document.write( "\n" ); document.write( "how fast would the car have been traveling for a braking distance of 150feet? round to nearest mile per hour.
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Algebra.Com's Answer #238118 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
The braking distance (in feet) of a car going V mph is given by
\n" ); document.write( " \"d%28v%29=v%5E2%2F20%2Bv\" v is greater or equal to 0.
\n" ); document.write( "how fast would the car have been traveling for a braking distance of 150feet?
\n" ); document.write( " round to nearest mile per hour.
\n" ); document.write( ":
\n" ); document.write( "Write it:
\n" ); document.write( "\"v%5E2%2F20%2Bv\" = 150
\n" ); document.write( "multiply by 20, results
\n" ); document.write( "v^2 + 20v = 20(150)
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\n" ); document.write( "v^2 + 20v = 3000
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\n" ); document.write( "v^2 + 20v - 3000 = 0
\n" ); document.write( "Solve for v using the quadratic formula
\n" ); document.write( "\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"
\n" ); document.write( "in this equation; x=v; a=1; b=20; c= -3000
\n" ); document.write( "\"v+=+%28-20+%2B-+sqrt%2820%5E2-4%2A1%2A-3000+%29%29%2F%282%2A1%29+\"
\n" ); document.write( ":
\n" ); document.write( "\"v+=+%28-20+%2B-+sqrt%28400-%28-12000%29+%29%29%2F2+\"
\n" ); document.write( ":
\n" ); document.write( "\"v+=+%28-20+%2B-+sqrt%2812400+%29%29%2F2+\"
\n" ); document.write( "Two solutions, we only want the positive solution
\n" ); document.write( "\"v+=+%28-20+%2B+111.355%29%2F2+\"
\n" ); document.write( "v = \"91.355%2F2\"
\n" ); document.write( "v = 45.68 mph for a stopping distance of 150 ft
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\n" ); document.write( "See if that flies in the original equation
\n" ); document.write( "\"d%28v%29=45.68%5E2%2F20%2B45.68\"
\n" ); document.write( "\"d%28v%29=2086.45%2F20%2B45.68\"
\n" ); document.write( "\"d%28v%29=104.32%2B45.68\"
\n" ); document.write( "d(v) = 150.00, confirms our solution
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