Where does the force actual act? 
Why are the two mass of $m_1$ and $m_2$ not multiplied by minus one?
I know that two minus multiplied gives you plus by but I mean the two masses are attracting so they should have a sign like so $$F_g=\frac{G(-m_1) \times (-m_2)}{r^2}$$
I ask these because equation would actual give the right explanation as to what the equation is actual doing, meaning that the two mass are attracting.
 A: Mass is always positive, that's why.
The equation you linked to is correct.  Your equation is also correct, although the minus signs are unconventional and they will always cancel.  Why would you include those minus signs?   But your equation is for the magnitude of the force, and says nothing about direction.  The equation you linked to is a vector equation, and includes both magnitude and direction.
Also ... please edit your post to include the actual image, rather than a link to it.   The link points only to a small image, so there is no problem including it in the question.
A: 
Actually the picture you've attached is a little sketchy about which force is acting on what. If I denote the force on $2$ due to $1$ as $F_{21}$ then the force on two would be given as in the picture. What the negative sign means is that the force on $2$ due to $1$ acts towards the direction of -$\vec{r}$  (as the force is always attractive). The vector expression for the force $F_{21}$ includes the minus sign primarily because the direction of the force on $2$ due to $1$ is in a sense opposite to the way $\vec{r}$ is defined. You might need to visit vectors for more clarity (if you haven't already).
A: In reality the force acts on all parts of $m_1$ and $m_2$ but it varies due to the variation in distance for different areas of $m_1$ and $m_2$. In practice though, if $m_1$ and $m_2$ are spherical, the force can be assumed to act at the center-of-mass of each.
A: 
but I mean the two masses are attracting so they should have a sign like so $F_g=\frac{G(-m_1) \times (-m_2)}{r^2}$

I am not following the logic. Attraction implies reduction in distance, and not reduction in mass. Therefore the distance related quantities should contain the negative and not the mass ones. Maybe if you group the equation as
$$ \vec{F} = \left( \frac{ G m_1 m_2}{\| r \|^2}  \right) \left(- \frac{ \vec{r} }{ \| \vec{r} \| }\right) $$
it would make more sense to you. The first part is the magnitude and the second part the direction.
