Quadratic A

$x^2 + 2ax + 10 - 3a > 0$

Since , this expression always remains positive, the parabola of this expression does not intersect X axis. Thus , the discriminant of this expression is $less$ than zero.

$\Rightarrow (2a)^2 - 4(1)(10 - 3a) < 0$

$\Rightarrow 4a^2 - 4(10 - 3a) < 0$

$\Rightarrow 4a^2 - 40 + 12a) < 0$

$\Rightarrow 4a^2 + 12a - 40) < 0$

$\Rightarrow (4a - 8)(a + 5) < 0$

When $a = 2 , -5$ , this expression becomes $0$ .

Thus range of $a$ is $-5 < a < 2$

discriminant of the quadratic equation is less than zero, then we get it

Another simple solution obtained from the given options(only for quick solving.
(Not much recommended).

If the value of a=0, then the given condition is satisfied.Only option two includes zero in the interval..................hehe

$(x+a)^2+10-3a-a^2$ >0

$(x+a)^2$ > $(a-2)(a+5)$

For all values LHS $(x+a)^2$ >=0

so $(a-2)(a+5)$ <0

so -5<a<2