If set of values of a for which $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^2-(3+2\mathrm{a})\mathrm{x}+6, \mathrm{a}\ne 0$is positive for exactly three distinct negative integral values of x is (c, d], then the value of (c + d) is equal to__.

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^2-(3+2\mathrm{a})\mathrm{x}+6\\ =(\mathrm{ax}-3)(\mathrm{x}-2)\end{array}$
 

Here, roots of the equation
f(x) = 0 are 2 and 3/a, and f(0) = 6.
f(x) should be positive for exactly three negative integral values of x.
Therefore, graph of f(x) must be a downward parabola passing through $\mathrm{x}=2\text{ and }\mathrm{x}=3/\mathrm{a}\text{ and }-4\le \frac{3}{\mathrm{a}}<-3$
$\therefore \mathrm{a}\in \left(-1,-\frac{3}{4}\right]$
$\begin{array}{r}\therefore \mathrm{c}=-1,\mathrm{d}=-\frac{3}{4}\\ \Rightarrow \mathrm{c}+\mathrm{d}=-1.75\end{array}$