Number of integers satisfying 8+2x−x2>6−3x is

8+2x−x2>6−3x We must have 8+2x−x2≥0⇒ x∈[−2,4]----(1) For 6−3x<0 or x>2 , inequality is satisfied.  Let 6−3x≥0⇒ x≤2----(2)⇒ 8+2x−x2>36+9x2−36x⇒ 5x2−19x+14<0⇒ (5x−14)(x−1)<0⇒x∈1,145---(3) From (ii) and (iii), x∈(1,2]∴ finally x∈(1,4]