Question

how can I factor (3n-2)(4n+1)=0
I already know that I have to find a,b,c but theres no sign in between those two parentheses, what do I do?

Answer 1

It is factored. Are you trying to solve for n?

Answer 2

maybe he is trying to multiply them to put it back into the form:
\[an^2+bn+c\]
?

Answer 3

yes

Answer 4

Yes to which one?

Answer 5

Im trying to solve it in quadratic formula way

Answer 6

but theres no addition or sub. in between (3n-2)(4n+1)

Answer 7

If you are looking to use the quadratic formula then you would have to multiply the two binomials first. It would be faster to solve for n without doing so.

Answer 8

i think you are making the problem a little harder than it has to be. Unless the teacher specifically asks for you to use the quadratic formula, if it looks like the equation above i wouldnt do it that way.

Answer 9

hmmm ok so how exactly would you answer this question because I got 2/3, 1/4 but how they got it I dont know

Answer 10

You can solve for n by taking each binomial and setting it equal to zero. Then solving for n.

For example:
3n-2 = 0
4n +1 =0

Solve those for n and you will have the possible solutions.

Answer 11

k thanks and how about n2= -18-9n

Answer 12

and -2v2-v+12=-3v2+6v step by step please

Answer 13

if that is \[n ^{2} =-18-9n\]
then you would need to have all the variables on one side and you could use the quadratic formula.

Answer 14

on one side? like -18-9n=n2?

Answer 15

all on one side like
\[n ^{2}+9n+18=0\]

Answer 16

does it automatically become a plus ?

Answer 17

?

Answer 18

it became positive because we added it to both sides of the equation.