Where's also an affordable place to fix brakes? How's Midas and Goodyear?
2007-01-13 14:27:35 · 8 answers · asked by dimples 1 in Cars & Transportation ➔ Maintenance & Repairs
I would say to check with some of your friends and see if they want to change your brakes for you as it would save you alot of money. If you don't have anyone willing to do the job for you then I would recommend Pep Boys, Midas, or Goodyear as they all are fairly resonable on pricing (as mechanics go) and will usually have it all done in less than two hours. I highly advise against taking to the dealership as they'll not only charge you an arm and a leg for it, but will take most of the day in repairing your vehicle.
2007-01-13 14:34:00 · answer #1 · answered by Strikernow 4 · 1⤊ 0⤋
I got mine done at Firestone, and they used high quality pad and they had the lowest prices around, even over Goodyear. They've done all types of work on my 2002 GMC Envoy and have never had a problem. I'd recommend them to anyone.
Check their website, they usually have some really good and usable promotions.
2007-01-13 23:06:59 · answer #2 · answered by Ariel G 1 · 0⤊ 0⤋
goodyear is good
try tire discouters also
or walmart
2007-01-17 18:20:53 · answer #3 · answered by peterosefan1414 3 · 0⤊ 0⤋
Midas is the best, i get it done there
2007-01-13 22:30:53 · answer #4 · answered by jdog33 4 · 0⤊ 1⤋
an independent shop. stay away from midas and jiffy lube.
2007-01-13 22:30:33 · answer #5 · answered by Anonymous · 3⤊ 0⤋
get some guy to do it for you. they are not hard to change and you will save the money. i can do them in 30 minutes but i dont know you, lol.
2007-01-13 22:31:56 · answer #6 · answered by Anonymous · 0⤊ 0⤋
talk with others see were there rec first then get more then one estimate before have in work performed
2007-01-13 22:32:19 · answer #7 · answered by Clint S 1 · 0⤊ 0⤋
dont waste your money just find someone you know to do it for you. because it's so easy a caveman can do it
2007-01-13 22:34:56 · answer #8 · answered by delicious buffet of manliness 2 · 0⤊ 0⤋