If x2+2ax+10−3a>0 for all x ∈ R, then
-5 < a < 2
a < -5
a > 5
2 < a < 5
According to given condition,
4a2−4(10−3a)<0 ⇒ a2+3a−10<0
⇒ (a+5)(a−2)<0⇒−5<a<2.