$x^2 - (m - 2)x + 6 = 0$ : Possible values of $m$. 
One of the roots of the equation $x^2 - (m - 2)x + 6 = 0$ lies between
$(0, 2)$ and the other root lies between $(3, 4)$. It is known that
$m$ is an integer. What is the sum of all the possible values of $m$?

Let $a$ be the root that lies between $(0, 2)$ and let $b$ be the root that lies between $(3, 4)$. I can write it mathematically as
$0 \lt a \lt2$ and $3 \lt b \lt4$ and also we can say that $3 \lt a+b \lt6$.
$a+b=m-2$ and this means that $3 \lt m-2 \lt6 \Rightarrow 5 \lt m \lt 8$. So the possible integer values of $m$ should be $6$ and $7$ and the possible sum should be $13$.
But as per the solution this is not the correct answer. What have I done wrong? Please help !!!
Thanks in advance !!!

The problem with using Viete's relations the way you did for the inequality is that they apply even when the roots are complex.  So $ \ x^2 - (6-2)x + 6 \ = \ 0 \ $ has the roots $ \ 2 \ \pm \ i\sqrt2 \ \ , $ which sum to $ \ 4 \ \  \ , $ satisfying $ \ 3 \ < \ a+b \ < \ 6 \ \ . $
One way we can look at the quadratic polynomial is to consider its "vertex form" describing an "upward-opening" parabola with $ \ y-$intercept $ \ (0 \ , \ 6 ) \ \ . $  "Completing the square" produces
$$ x^2 \ - \ (m-2)x \ + \ 6 \ \ = \ \ \left(x \ - \ \frac{m-2}{2} \right)^2 \ + \ \left(6 \ - \ \frac{(m-2)^2}{4} \right) \ \ , $$
which tells us that the vertex is the $ \ x-$intercept of the parabola for
$$  6 \ - \ \frac{(m-2)^2}{4}  \ \ = \ \ 0 \ \ \Rightarrow \ \ m \ \ = \ \ 2 \ \pm \ 2·\sqrt6 \ \ . $$
Our quadratic equation then has real roots for $ \   m \  <   \ 2 \ - \ 2·\sqrt6 \   $ and $ \   m \  >   \ 2 \ + \ 2·\sqrt6 \ \ .  $ But the $ \ y-$intercept at $ \ (0 \ ,  \ 6 ) $ means that we cannot has positive real roots for $ \   m \  <   \ 2 \ - \ 2·\sqrt6 \  \ ,  $ since the vertex is "to the left" of the $ \ y-$axis, leaving us with $ \   m \  >   \  \approx \ 6.899 \ \ .  $
With $ \ m \ $ required to be an integer, we can assess cases directly.  For $ \ m \ = \ 7 \ \ , $ the equation is $ \ x^2 - (7-2)x + 6 \ = \ (x-2)·(x-3) \ = \ 0 \ \ ; $ unfortunately, neither of the roots are in the specified open intervals.  The next integer case leads to  $ \ x^2 - (8-2)x + 6 \ = \  0 \ \ , $ with roots $ \ 3 - \sqrt3 \ \approx \ 1.27 \ $ and $ \ 3 + \sqrt3 \ \approx \ 4.73 \ \ , $ the larger of these being "to the right" of the interval $ \ (3 \ , \ 4 ) \ \ . $  Using larger integers simply shifts the vertex "lower and to the right", making the larger of the roots even larger.  Thus, we  confirm Z Ahmed's conclusion that there are no possible integer values of $ \ m \ $ to sum.  (Trick question...)
