This problem can be done without needless casework. (For negative values of $x$, the left hand side will be negative, so we only need to consider positive values of $x$.) The key observation is that for $x \in[2,3), 122$ is an extremely large value for the expression. Indeed, we observe that: $\lfloor x\rfloor =2 \lfloor x\lfloor x\rfloor\rfloor \leq 2(3)-1 =5 \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor \leq 3(5)-1 =14 \lfloor x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\rfloor \leq 3(14)-1 =41 x\lfloor x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\rfloor <3(41) =123$. So the expression can only be as large as 122 if ALL of those equalities hold (the the fourth line equaling 40 isn't good enough), and $x=\frac{122}{41}$. Note that this value is extremely close to 3. We may check that this value of $x$ indeed works. Note that the expression is strictly increasing in $x$, so $x=\frac{122}{41}$ is the only value that works.