We have the quadratic equation $64x^2 + 5Nx + 1 = 0$. For it to have no real roots, the discriminant ($b^2 - 4ac$) should be less than 0. Here, $a = 64$, $b = 5N$, and $c = 1$.

This gives us :

$(5N)^2 - 4\times64\times1 < 0$

$\Rightarrow 25N^2 < 256$

$\Rightarrow N^2 < \frac{256}{25}$

$\Rightarrow N < \sqrt{\frac{256}{25}} = \frac{16}{5}$

Since $N$ must be an integer (as it represents the number of tosses), the possible values of $N$ are 1, 2, or 3.

The probability of getting the first head on the $n$-th toss (given the probability of getting a head is $1/4$) is given by the geometric distribution formula, $(1 - p)^{n-1}\times p$.

So, the probability for our specific values of $N$ is:

$P(N=1) = (1 - 1/4)^{1-1}\times(1/4) = 1/4$

$P(N=2) = (1 - 1/4)^{2-1}\times(1/4) = 3/4 \times 1/4 = 3/16$

$P(N=3) = (1 - 1/4)^{3-1}\times(1/4) = (3/4)^2 \times 1/4 = 9/64$

Therefore, the total probability (p/q) is :

$p/q = P(N=1) + P(N=2) + P(N=3)$

$= 1/4 + 3/16 + 9/64$

$= 16/64 + 12/64 + 9/64$

$= 37/64$

So, $p = 37$, $q = 64$ and $q-p = 64 - 37 = 27$.

Therefore, $q-p$ is equal to $27$.