Can someone help me with this Algebra problem?

[OCNewbie]

Final is tomorrow and this is on the practice test

(7a²-3a)-[(2a²-4a)+(3a²+6)]

I presume there is factoring involved, but I'm unsure of the answer. I know you have to combine like terms, etc., but I'm just stuck.

btw, to do the X² power of 2 symbol, hold down alt and type 253 on numpad.

TYIA for any assistance =)

 

[txrandom]

There's nothing to solve, that's not an equation.

 

[hypn0tik]

(7a²-3a)-[(2a²-4a)+(3a²+6)] = 7a^2 - 3a -2a^2 + 4a -3a^2 -6 = (7-2-3)a^2 + (-3 + 4)a -6 = 2a^2 +a -6 = (2a - 3)(a + 2)

Edit: a^2 = a²

 

[txrandom]

2a^2 + a - 6 = 0

(2a - 3) * (a + 2) = 0

a= 3/2 , -2

 

[OCNewbie]

Yeah, I think I figured it out. I think I was just thrown off by all the parenthesis/brackets. I guess there is no multiplication involved, so it's just a straight transfer the negative, after combining the terms in the brackets, etc. That way gave me the right answer. I have an answer sheet, just not a how-to-do-the-problem listing.

I think I was making it much more complicated than it really was. Thanks again =)

Edit - Answer according to answer sheet - (2a² + a - 6)

Don't have to set equal to zero, that's the final answer right there in parenthesis.