If set of values of a for which f(x)=ax2−(3+2a)x+6, a≠0is positive for exactly three distinct negative integral values of x is (c, d], then the value of (c + d) is equal to__.

f(x)=ax2−(3+2a)x+6      =(ax−3)(x−2) Here, roots of the equationf(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 x=2 and x=3/a and −4≤3a<−3∴ a∈−1,−34∴ c=−1,d=−34⇒ c+d=−1.75