$(x^2+2mx+7m-12)(4x^2-4mx+5m-6)=0$ have two distinct real roots Find the range of $m$ so that the equation $(x^2+2mx+7m-12)(4x^2-4mx+5m-6)=0$ have two distinct real roots.
Either $(x^2+2mx+7m-12)=0$ or $(4x^2-4mx+5m-6)=0$
The discriminant of $(x^2+2mx+7m-12)=0$ is $4(m-3)(m-4)$ 
and the discriminant of $(4x^2-4mx+5m-6)=0$ is $16(m-2)(m-3)$
I don't know how to do it further.

We can also look at the "vertex form" of each quadratic factor of $ \  (x^2+2mx+7m-12)·(4x^2-4mx+5m-6) \ $ to see what they contribute to their product function.
The first factor can be written as
$$ f(x) \ \ = \ \ x^2 \ + \ 2mx \ + \ (7m - 12) \ \ = \ \ (x \ + \ m)^2 \ + \ (7m - 12) \ - \ m^2  $$ $$ = \ \ (x \ + \ m)^2 \ - \ (m^2 \ - \ 7m \ + \ 12)  \ \ = \ \ (x \ + \ m)^2 \ - \ [ \ (m \ - \ 3)·(m \ - \ 4) \ ] \ \   $$
[as expected, the factors of the discriminant you found appear in the bracketed term of the "completed square"].  The vertex of the "upward-opening" parabola corresponding to this factor is  then "above" the $ \ x-$axis when the term in brackets is negative, which occurs for $ \ 3 < m < 4 \ \ . $  We can then say that this factor has
•  no real zeroes for $ \ 3 \ < \ m \ < \ 4 \ \ , $
•  one real zero at $ \ x \ = \ -m \ $ when $ \ m \ = \ 3 \ $ or $ \ m \ = \ 4 \ \ , \ \ $ and
•  two real zeroes for $ \ m \ < \ 3 \ $ or $ \ m \ > \ 4 \ \ . $
Considering the second factor in this way, we find
$$ g(x) \ \ = \ \ 4x^2 \ - \ 4mx \ + \ (5m - 6) \ \ = \ \ 4·\left(x \ - \ \frac{m}{2} \right)^2 \ + \ (5m - 6) \ - \ 4·\left(\frac{m}{2} \right)^2  $$ $$ = \ \ 4·\left(x \ - \ \frac{m}{2} \right)^2 \ - \ (m^2 \ - \ 5m \ + \ 6)  \ \ = \ \ 4·\left(x \ - \ \frac{m}{2} \right)^2 \ - \ [ \ (m \ - \ 2)·(m \ - \ 3) \ ] \ \  . $$
The vertex for the "upward-opening" parabola here is again "above" the $ \ x-$axis when the term in brackets is negative for $ \ 2 < m < 3 \ \ . $  So we conclude that this factor has
•  no real zeroes for $ \ 2 \ < \ m \ < \ 3 \ \ , $
•  one real zero at $ \ x \ = \ \frac{m}{2} \ $ when $ \ m \ = \ 2 \ $ or $ \ m \ = \ 3 \ \ , \ \ $ and
•  two real zeroes for $ \ m \ < \ 2 \ $ or $ \ m \ > \ 3 \ \ . $
The quartic polynomial that is the product of these two factors has the number of real zeroes that is the sum of the number of real zeroes in the two factors.  Therefore, $ \ f(x) \ · \ g(x) \ $ has, for varying $ \ m \ \ : $
$ \mathbf{m \ < \ 2 \ \ : } \ \ \ 2 + 2  \ = \ 4 \ $ real zeroes;
$ \ \ \mathbf{m \ = \ 2 \ \ : } \ \ \ 2 + 1  \ = \ 3 \ $ real zeroes (one of them at $ \ x \ = \ \frac{m}{2} \ = \ \frac{2}{2} \ = \ +1 \ \ ) \ ; $
$ \mathbf{2 \ < \ m \ < \ 3 \ \ : } \ \ \ 2 + 0  \ = \ 2 \ $ real zeroes;
$ \ \ \mathbf{m \ = \ 3 \ \ : } \ \ \ 1 + 1  \ = \ 2 \ $ real zeroes (one of them at $ \ x \ = \ -m \ = \ -3  \ \ , $ the other at $ \ x \ = \ \frac{m}{2} \ = \ +\frac{3}{2} \ \ ) \ ; $
$ \mathbf{3 \ < \ m \ < \ 4 \ \ : } \ \ \ 0 + 2  \ = \ 2 \ $ real zeroes;
$ \ \ \mathbf{m \ = \ 4 \ \ : } \ \ \ 1 + 2  \ = \ 3 \ $ real zeroes (one of them at $ \ x \ = \ -m \ =  \ = \ -4 \ \ ) \ ; $
and  $ \ \ \mathbf{ m \ > \ 4 \ \ : } \ \ \ 2 + 2  \ = \ 4 \ $ real zeroes.
Hence, the quartic polynomial has two distinct real zeroes for $ \ \mathbf{2 \ < \ m \ < \ 4} \ \ ; \ \ $ these are the zeroes of either factor alone or from both vertices at once $ \ ( \ m \ = \ 3 \ ) \ \ . $
The polynomial never has less than two real zeroes for any value of $ \ m \ \ . \ $ The third zero for $ \ m \ = \ 2 \ $ and $ \ m \ = \ 4 \ $ must be a "double zero"; the two zeros for $ \ m \ = \ 3 \ $ are both "double zeroes".  In fact, this polynomial never has  complex zeroes.
[There are places where one could be deceived by using a low-resolution graph or not examining a graph carefully enough.  The function $ \ f(x) \ · \ g(x) \ $ has a turning point extremely close to the $ \ x-$axis for
$ m \ \approx \ -2.25 \ $ near $ \ x \ \approx \ -3.482 \ \ $ (slightly below); $ \ m \ \approx \ +1.12 \ $ near $ \ x \ \approx \ +1.205 \ \ $ (slightly below); and $ \ m \ \approx \ +1.53 \ $ near $ \ x \ \approx \ +0.362 \ \ $ (slightly above). ]
