For all $$'x'\,,{x^2} + 2ax + 10 - 3a > 0,$$      then the interval in which $$'a'$$ lies is

Solution :
$$\eqalign{ & {\text{Given equation - }} \cr & {x^2} + 2ax + \left( {10 - 3a} \right) > 0{\text{ for all }}x \in R \cr & {\text{Here, }}A = 1,\,B = 2a,\,C = \left( {10 - 3a} \right) \cr & {\text{As we know that,}} \cr & A{x^2} + Bx + C > 0{\text{ for all }}x \in R{\text{ if}} \cr & A > 0{\text{ and }}D < 0 \cr & A = 1 > 0 \cr & {\text{Now, }}D < 0 \cr & {B^2} - 4AC < 0 \cr & {\left( {2a} \right)^2} - 4\left( 1 \right)\left( {10 - 3a} \right) < 0 \cr & 4{a^2} - 40 + 12a < 0 \cr & {a^2} + 3a - 10 < 0 \cr & \left( {a + 5} \right)\left( {a - 2} \right) < 0 \cr & a \in \left( { - 5,\,2} \right) \cr & \therefore \, - 5 < a < 2 \cr} $$