Let the base be n.

Then, the given equation is:

X^2-(n+7)x+(7n+3)=0 .

By the given conditions, 

(n+7)^2 - 4(7n+3) = t^2, when t is an integer

Thus, we have:

n^2-14n+37=t^2

-> (n-7)^2 - t^2 =12

-> the expression on the lhs factorizes as: (n-7+t)(n-7-t). Since n-7+t and n-7 -t must have the same parity, it follows that only 2*6, (-6)*(-2) are the valid integer factorizations of it

Or, (n-7+t)(n-7-t) = (2)*(6), (6)*(2), (-6)*(-2), (-2, -6)

Or,  (n-7, t) = (4, +/-2), (-4, +/-2)

Or,  (n,t) = (11, +/-2), (3, +/-2) are purportedly  the only possible integer pairs.

However, neither 17 nor 73 is a valid number in base 3.

Consequently,  11 is the only possible  integer base satisfying the given conditions.

VERIFICATION 

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WHEN  the base is 11, then the given equation reduces to:

x^2-18x+80=0 in base ten

Then, (x-10)(x-8)=0

-> x=10,8

-> x = a, and 8 are the roots of the equation in base 11.