As 0 ≤ {x} < 1 then $2x \leq 6\{x\}^2 + \{x\} + 2x < 8 + 2x$. Hence 2x ≤ 2018 < 8 + 2x, from where we get that 1005 < x ≤ 1009. From these inequalities it follows that 1005 ≤ [x] ≤ 1009. By using the equation x = [x] + {x} we can write the inequality in the following way: $6\{x\}^3 + \{x\}^2 + 3\{x\} = 2018 – 2[x].$ $f (t) = 6t^3 + t^2 + 3t$ is an increasing function as $f’(t) = 18t^2 + 2t + 3 > 0$ and we also have that f (0)= 0 and f (1) = 10. Therefore when t ∈ [0, 1) the function takes all values in the interval [0, 10) exactly once. When [x] = 1005, 1006, 1007, 1008, 1009 we have that 2018 – 2[x] = 8, 6, 4, 2, 0. Hence the equation has 5 solutions.