Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

Solution for $x$ is $x\in (-45,-\frac{2024}{45}]$.

Proof:

Let $[x] = \lfloor x\rfloor$ and $\{x\} = x - [x] \implies [x] = x-\{x\}$

$\implies 2025 > [x]^2+[x]\{x\} \geq 2024$

Case 1: 

$[x]$ is positive.

$[x]$ is an integer and, $0\leq \{x\}<1 \implies 0\leq \{x\}[x]<[x]$

[[x[x]] = 2024 \implies [[x]^2+\{x\}[x]] = 2024\]

From the first part,

\[2025 >[x]^2+[x]\{x\}\implies 2025 > [x]^2 \implies [x]<45\]

From the second part,

\[[x]^2+[x]\{x\} \geq 2024\implies [x]^2 + [x] \geq 2024 \implies [x]>44\]

So, $45>[x]>44$, no integer solution for $[x]$

Case 2:

 $[x]$ is negative.

$[x]$ is an integer and $0\leq \{x\}<1 \implies 0\geq \{x\}[x]>[x]$

\[[x[x]] = 2024 \implies [[x]^2+\{x\}[x]] = 2024\]

From the first part,

\[2025 >[x]^2+[x]\{x\}\implies 2025 > [x]^2+[x] \implies [x]>-46\]

From the second part,

\[[x]^2+[x]\{x\} \geq 2024\implies [x]^2 \geq 2024 \implies [x]<-44\]

So, $-44>[x]>-46$ and so $[x] = -45$.

So, \[[-45x]=2024 \implies 2025 > -45x \geq 2024 \implies -\frac{2024}{45} \geq x > -45\]

So, the only solutions for $x$ is: $x\in (-45,-\frac{2024}{45}]$