SOLUTION: please factor completely 4c^2+9-12c What do I do first?

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Question 346178: please factor completely
4c^2+9-12c
What do I do first?
Found 3 solutions by Fombitz, rfer, ewatrrr:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
4c%5E2-12c%2B9=0
Oops!
You're correct that is the answer.
Sorry for screwing you up there.
(2c-3)(2c-3)=4c^2-6c-6c+9=4c^2-12c+9
I will correct it on the website too.
I could say I was just testing you but I just screwed up.
Good work!
highlight%284c%5E2-12c%2B9=%282c-3%29%5E2%29

Answer by rfer(16322) About Me  (Show Source):
You can put this solution on YOUR website!
4c^2-12c+9
(2c- 3)(2c- 3)

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
Hi,
4c%5E2%2B9-12c
.
First put in this form:
4c%5E2-12c%2B9
.
Check for factoring: *Hint: First Ask Yourself: What numbers added together have a sum of -12 (-6 and -6)
.
Then:
What two numbers have a product of 4 and what two numbers have a product of 9 such that the product of these numbers will be -6
.
highlight%28%282c-3%29%282c-3%29%29
.
Check you answer:
%282c-3%29%282c-3%29 = 4c%5E2-12c%2B9