if you have a number x then the consecutive number is x+1
the multiplication of two consecutive numbers can then be written as x*(x+1)=...
if you find an integer solution for this equation you can write the number as two consecutive integers multiplied with each other, if not you found the exception you are looking for
E)
x(x+1)=9900
x²+x=9900
-> complete square
x²+x+(1/2)²=9900+(1/2)²
(x+(1/2))²=9900+(1/4)
x+(1/2)=sqrt(9900+(1/4))
x=sqrt(9900+(1/4))-(1/2)
x=99
->9900=99*(99+1)=99*100
if we replace 9900 in the last lines with the other numbers we can check them quickly too:
F)
x=sqrt(15750+(1/4))-(1/2)
x=125
->15750=125*(125+1)=125*126
G)
x=sqrt(20448+(1/4))-(1/2)
x=142.497...
->not an integer
->can not be written as two consecutive numbers multiplied
H)
x=sqrt(38612+(1/4))-(1/2)
x=196
->38612=196*(196+1)=196*197
so G is the only number which can not be written as two consecutive integers
