#### Flo Rida

##### Junior Member
I haven't had trouble until I got to this question:

2x2+3x+6=4x

i subtract 4x and i get 2x2-x+6=0
then i multiply a term with c term and I get x2-x+12=0.

i dont know any factors of 12 that add to -1. Are there any?

thanks
flo

#### lev888

##### Elite Member
then i multiply a term with c term and I get x2-x+12=0.

#### Flo Rida

##### Junior Member
My teacher taught me to multiply a by c to get x2 alone
ex:
2x2-5x+2=0 becomes x2-5x+4.
now, it would be (x-1)(x-4) and then i have to divide by the a term
the answer would be (2x-1)=0 and (x-2)=0,

problem with this question above is that i cant find factors that multiply to 12 and add to -1

#### lev888

##### Elite Member
My teacher taught me to multiply a by c to get x2 alone
ex:
2x2-5x+2=0 becomes x2-5x+4.
now, it would be (x-1)(x-4) and then i have to divide by the a term
the answer would be (2x-1)=0 and (x-2)=0,

problem with this question above is that i cant find factors that multiply to 12 and add to -1
Again, step by step, how did 2x2-5x+2=0 become x2-5x+4=0?

#### Flo Rida

##### Junior Member
Again, step by step, how did 2x2-5x+2=0 become x2-5x+4=0?
2 is removed from the a-term (x2) and distributed to the c-term (2), which becomes 4.

#### lev888

##### Elite Member
2 is removed from the a-term (x2) and distributed to the c-term (2), which becomes 4.
It's not clear to me what you are trying to do. But this approach certainly changes the equation. Do you see it? You can manipulate equations _only_ by applying the same operations to both sides. If you want to make the first coefficient 1 you need to divide both sides by 2. Maybe this is the method your teacher used?

#### Flo Rida

##### Junior Member
It's not clear to me what you are trying to do. But this approach certainly changes the equation. Do you see it? You can manipulate equations _only_ by applying the same operations to both sides. If you want to make the first coefficient 1 you need to divide both sides by 2. Maybe this is the method your teacher used?
No, she's been using this, and it worked for all the others, but I really don't understand why it isn't working for this.

#### lev888

##### Elite Member
Can you post a screen cap of the problem? And examples of your teacher's method if available.

#### Flo Rida

##### Junior Member
Can you post a screen cap of the problem? And examples of your teacher's method if available.
Ok give me a few minutes

#### Flo Rida

##### Junior Member
Original question:
Solve the equations by factoring.
2x2+3x+6=4x

#### Flo Rida

##### Junior Member
Example w/other question (#6 only)

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#### lev888

##### Elite Member
Original question:
Solve the equations by factoring.
2x2+3x+6=4x
wolframalpha.com says this equation has only complex roots, that's why I suspect you have a typo.

#### lev888

##### Elite Member
Example w/other question (#6 only)
This is ridiculous. The answer is correct, but I have no idea what this method is and why we need to make the solution more complicated than necessary.

2x^2-32 = 0
Divide both sides by 2:
x^2-16 = 0
(x-4)(x+4) = 0
x = -4, 4

#### Flo Rida

##### Junior Member
wolframalpha.com says this equation has only complex roots, that's why I suspect you have a typo.
Ok, well I guess that clears it up.
This is ridiculous. The answer is correct, but I have no idea what this method is and why we need to make the solution more complicated than necessary.

2x^2-32 = 0
Divide both sides by 2:
x^2-16 = 0
(x-4)(x+4) = 0
x = -4, 4
thank you so much! this helps with the other ones.

#### lev888

##### Elite Member
How did you solve the original problem?

#### Flo Rida

##### Junior Member
How did you solve the original problem?
I checked my calculator and it does involve roots. i asked my math teacher and she removed the question and said it was a typo

#### ghanasur

##### New member
i tried solving it but unfortunately i was unable to solve it using factoring

#### JeffM

##### Elite Member
i tried solving it but unfortunately i was unable to solve it using factoring
First, this thread is months old.

Second, every quadratic can be factored using complex numbers although not all quadratics can be factored using real numbers.

Third, as was discussed at length in the thread, this quadratic is one of those that cannot be factored using real numbers.