A quadratic question

So, two cases: if b is 0, we must have ac = 0. Since a isn’t 0,
c must be, so that’s 200 equations we pick up there.
Otherwise, b isn’t zero. Certainly b has to be even. So let b
= 2k, where k runs from -50 to 50, skipping 0. Obviously, the
sign isn’t going to matter here, so we can just take k running
from 1 to 50 and double our answer at the end. Then we have k^2
= ac. So, we need the number of factorizations of 1^2, 2^2,
3^2, …, 50^2 into positive integers. Then we double that
(because we can take negatives of both a and c) and then we
double the result again (because I said so earlier, for the sign
of b) and add the 200 in from the other case. Now, a
factorization into two parts is the same as a factor, so we need
the sum of the number of factors of the first 50 squares.
Unfortunately, I have no clever way of approaching that problem
at all. Mathematica tells me that the first fifty squares have
the following numbers of factors: {1, 3, 3, 5, 3, 9, 3, 7, 5, 9,
3, 15, 3, 9, 9, 9, 3, 15, 3, 15, 9, 9, 3, 21, 5, 9, 7, 15, 3,
27, 3, 11, 9, 9, 9, 25, 3, 9, 9, 21, 3, 27, 3, 15, 15, 9, 3, 27,
5, 15}, for a total of 470, which means that 2080 of the total
8080200 such equations have a double root, 0.0025…%.