How does one rewrite
$$ l = \left\lfloor\frac{10\lfloor\frac{d}{7}\rfloor+\lfloor\frac{d}{10}\rfloor(\lfloor\frac{d}{10}\rfloor+1)}{2}\right\rfloor $$
in terms of $l$ such that the isolate variable, $d$, represents the least possible value corresponding to an input $l$ for both $l$ and $d$ are positive integers?
Thanks to Ross Millikan's response, $$ d = \left\lfloor\frac{\sqrt{1070^2+4*7*1400l}-1070}{14}\right\rfloor $$ has been found to be a rough approximation of the inverse of the original equation. The original equation produces the following input-output pairs in the form $(d,l)$: $$ A(30,26), B(50,50), C(70,78), D(90,105), E(100,125) $$ For each of these pairs, the input is the least possible value of $d$ that is mapped to its respective output of $l$. However, if you input the set of values of $l$ into the inverted equation, the following coordinate pairs in the form $(d,l)$ are produced: $$ A'(28,26), B'(49,50), C'(70,78), D'(87,105), E'(99,125) $$ That is, the inverted equation produces impossibly low outputs. Notice that points $C$ and $C'$ remain intact. I believe that this is because $l=70$ for these two points: 70 is divisible by both $7$ and $10$.
Say that the original equation represents the number of leaves that a growing caterpillar has eaten since its birth at day $d=0$. The caterpillar sleeps when it is not eating. Once every seventh day after its birth, first occurring at $d=7$, it awakens to consume $5$ leaves, then returns to sleep. Every tenth day after its birth, first occurring at $d=10$, it consumes one leaf for every tenth day it has been alive. Thus, at $d=7$, $l=5$, at $d=10$, $l=6$, at $d=14$, $l=11$, at $d=20$, $l=13$, and so forth. In the original equation, the factors $\lfloor\frac{d}{7}\rfloor$ and $\lfloor\frac{d}{10}\rfloor$ are used as checks to ensure that the caterpillar eats no leaves while it is sleeping; if $d$ is not a multiple of $7$ or $10$, then the caterpillar eats not even a fraction of a leaf, for it is asleep.
So how can the original equation be inverted such that passing in a value of $l$ produces an the least possible output value $d$ for which the original equation would produce the lowest possible value of $l$ that is at least the value of $l$ that was passed in to the inverted equation to begin?
One approach is to start by ignoring all the floors, giving $1400l=570d+7d^2$ and use the quadratic equation to get $d=(-570+\sqrt {570^2+4\cdot 7 \cdot 1400l})/14$, where I took the plus sign because $d$ should be positive. This will give a lower bound for $d$. Now just count upward until you find what you like. For example, if $l=100$, my equation gives $d=106.4511$. Rounding up to $107$ and inserting the floors gives $l=92$. Counting up, we find $(d=109, l=92), (d=110,l=103)$ as solutions to your original equation. Which of these do you consider best?