Conventional Approach :
Given inequality is $|x+3|>x^2-11x+30$
We consider 2 cases 
Case (i) : $x>-3$
$|x+3|=x+3$
$x+3>x^2-11x+30\Rightarrow x^2-12x+27<0$
$(x-9)(x-3)<0$
 x= (3,9)
Case (ii) ; $x<-3$
|x+3|= -x-3
$-x-3>x^2-11x+30$
$\Rightarrow x^2-10x+33<0$
$\Rightarrow x^2-10x+25+8<0=(x-5{)}^2+8<0$
$(x-5{)}^2+8$ is always a positive quantity, irrespective of the value of x. Hence we do not have any value for x in the interval. Hence  the valid interval for x is (3,9). 

Shortcut- Elimination Approach

Step 1: Choose a value for x which is present in 2 answer options and absent in the other 2. Eg) x=4 satisfies the range specified in option (a) and (d), but does not form a part of the range in option (b) and (c)

At x=4

LHS= |4+3|=7

RHS= 16 - 44+30 = 2. LHS$>$RHS. It is satisfied. This means that answer options, which do not include x=4 can be directly eliminated as they can NEVER be the right answer

Step 2- Choose a value for x which is there in one of the remaining 2 options and absent in the other

Eg) Take x=10. This option satisfies the range in option (a) but is absent in option (d)

At x=10

LHS= |10+3|=13

RHS= 100-110+30= 20

LHS$<$RHS. This violates the condition in the question. This means that option (a) can never be the right answer.

Thus, by elimination, answer is option (d)