Struggling to solve an equation containing a floor function I am trying to solve the following equation ($x\in \mathbb R)$
$$
2\lfloor{(x+1)^2+8}\rfloor=(x+1)(2x+3) \tag{1}
$$
I have tried substituting $x$ with its split form (integer part ($n$) + decimal part ($\alpha$)), i.e., $x=n+\alpha$ with $n=\lfloor x \rfloor.$
The substitution yields:
$$
\lfloor 2\alpha^2 + 4\alpha n + 4\alpha\rfloor=2\alpha^2+4\alpha n+5\alpha + 3n -15\tag{2}
$$
or equivalently
$$
\lfloor 2\alpha(\alpha + 2n +2)\rfloor= \alpha(2\alpha + 4n + 5) + 3n -15\tag{3}
$$
and here I am stuck and I don't know how to proceed, or whether I've already complicated matters more than needed.  Any hints would be much appreciated.

Let $x$ be an integer.
Set $y = x+1$, then we have $2\lfloor y^{2}+8 \rfloor=y(2y+1)=2y^{2}+y. $
Since $\lfloor a+c \rfloor = \lfloor a \rfloor +c$ for all integers $a$, $2\lfloor y^{2}+8 \rfloor=2(\lfloor y^{2} \rfloor + 8)=2\lfloor y^{2} \rfloor +16=2y^{2}+y$ from above.
If $y > 16$, then $2\lfloor y^{2} \rfloor = 2y^{2} + c$ for some $c > 0 \implies 2y^{2}<2 \lfloor y^{2} \rfloor \implies y^{2}<\lfloor y^2 \rfloor$ which is a contradiction as $\lfloor a \rfloor \leq a$ for all $a$ by definition.
Hence, we must check the finite number of cases of the form $y \leq 16 \implies x \leq 15$.
Since $x$ is an integer ($\implies y$ is an integer), then $2y^{2}+16=2\lfloor y^{2} \rfloor +16=2y^{2}+y \implies$
$y = 16 \implies x = 15.$
Therefore $15$ is the only integer solution.
