The number of integral values of a, $a\in(6,100)$ for which the equation $[tan\;x]^{2}+tan\;x-a=0$ has real roots (where [.] denotes greatest integer function) is:

Let y = tan x. The equation is y^2 + y - a = 0. For real roots, the discriminant D = 1 - 4(1)(-a) = 1 + 4a >= 0, which is always true for a > 6. However, since y = tan x, y can be any real number. The equation y^2 + y - a = 0 has roots y = (-1 +/- sqrt(1 + 4a)) / 2. For tan x to have these values, the roots must be real, which they are. The constraint is on the floor function [tan x]. The problem likely implies [tan x] is a constant k, so k^2 + k - a = 0, meaning a = k^2 + k. For a in (6, 100), k^2 + k - a = 0 implies k is an integer. For k=3, a=12; k=4, a=20; k=5, a=30; k=6, a=42; k=7, a=56; k=8, a=72; k=9, a=90. There are 7 such values.