You can use the floor with the difference of the desired quantity and a sufficiently large integer. In particular here we know:
$$ v > \frac{\sqrt{v-4 \times N}-1}{4} $$
Therefore:
$$ L = \left\lceil \frac{\sqrt{v-4 \times N}-1}{4} \right\rceil = v - \left\lfloor v - \frac{\sqrt{v-4 \times N}-1}{4} \right\rfloor $$
'Nuff said? There are other ways to do it, but you have $v$ on hand and this is probably the fewest additional operations (and avoids any case logic). Certainly worth a comment in the code so you don't have a long moment of confusion months down the road...