How many integers satisfy (sqrt n- sqrt 8836)^2<1?

Dear Vivek,

its a very easy question,
sqrt 8836 = 94 , let sqrt n=x
the equation becomes...
(x-94)^2 < 1
(x-94)^2 - 1 < 0
(x-95)(x-93) < 0
hence  93<x<95, so (93)^2 < n < (95)^2
8649 < n < 9025
hence the number of integers are 9024-8650+1=375
Hence the number of Integers are 375 (ANS).

Regards