Can you solve this equation?

Let the LHS of the equation be $S(x)$ as follows:

$\begin{aligned} S(x) & = \lfloor \sqrt 1 \rfloor + \lfloor \sqrt 2 \rfloor + \lfloor \sqrt 3 \rfloor + \lfloor \sqrt 4 \rfloor + \cdots + \lfloor \sqrt {x^2-1} \rfloor \\ & = \underbrace{1 + 1 + 1}_{1(2^2-1^2)} + \underbrace{\lfloor \sqrt 4 \rfloor + 2 + \cdots + 2}_{2(3^2-2^2)} + \underbrace{\lfloor \sqrt 9 \rfloor + 3 + \cdots + 3}_{3(4^2-3^2)} + \lfloor \sqrt {16} \rfloor + \cdots + \underbrace{\lfloor \sqrt {(x-1)^2} \rfloor + \cdots + \lfloor \sqrt {x^2-1} \rfloor}_{(x-1)(x^2 - (x-1)^2)} \\ & = \sum_{k=1}^{x-1} k((k+1)^2-k^2) = \sum_{k=1}^{x-1} k(2k+1) = \sum_{k=1}^{x-1} (2k^2 + k) \\ & = \frac {x(x-1)(2x-1)}3 + \frac {x(x-1)}2 \\ & = \frac {x(x-1)(4x+1)}6 \end{aligned}$

For $S(x) = 34$ , we have:

$\begin{aligned} \frac {x(x-1)(4x+1)}6 & = 34 \\ x(x-1)(4x+1) & = 204 \\ 4x^3-3x^2-x - 204 & = 0 \\ (x-4)(4x^2+13x+51) & = 0 \\ \implies x & = \boxed 4 \end{aligned}$

Floor function of first three natural numbers gives 1. 3*1=3. Total = 3

Floor function of next five gives 2. 5*2=10. Total = 13

Floor function of next seven gives 3. 7*3=21. Total = 34

3+5+7=15.

X^2-1=15.

x = 4.