Quadratic and Floor

Let $a=\lfloor x \rfloor$ and $b=\{x\}$ . $(a+b)^2-8a+10=0$ $b^2+2ab+a^2-8a+10=0$ $b=\frac{-2a\pm\sqrt{32a-40}}{2}=-a\pm\sqrt{8a-10}$

Since $b$ is real thus $8a-10\geq0$ and $a$ is positive.

We define $b$ to be positive, so we reject the negative root since otherwise $b$ would be negative.

$0 \leq b<1$ $0\leq-a+\sqrt{8a-10}<1$ $0\leq-a+\sqrt{8a-10} \mbox{ and} -a+\sqrt{8a-10}<1$ $a^2-8a+10\leq0 \mbox{ and } a^2-6a+11>0$ $(a-\frac{8-\sqrt{24}}{2})(a-\frac{8+\sqrt{24}}{2})\leq0 \mbox{ and } (a-3)^2+2>0$

The first inequality yields $4-\sqrt{6}\leq a \leq 4+\sqrt{6}$ and the second always holds true. Thus we have $a=2,3,4,5,6$ and there are $\boxed{5}$ solutions for $x$ .

(Namely $x=\sqrt{6},\sqrt{14},\sqrt{22},\sqrt{30} \mbox{ or } \sqrt{38}$ )

Call $S = x^2 - 8\lfloor {x}\rfloor + 10$

If $\lfloor {x}\rfloor \leq 0 \Rightarrow$ $\color{#3D99F6}\text{S > 0}$ .

Therefore $\lfloor {x}\rfloor > 0$

Now , if $\lfloor {x}\rfloor \geq 8$

$\Rightarrow x \geq \lfloor {x}\rfloor \geq 8$

$\Rightarrow x^2 \geq \lfloor {x}\rfloor^{2} \geq 8\lfloor {x}\rfloor \Rightarrow$ $\color{#3D99F6}\text{S > 0}$

Therefore , $\boxed{1 \leq \lfloor {x}\rfloor \leq 7}$ .

Checking the cases yields $\boxed{x = \sqrt{6} , \sqrt{14} , \sqrt{22} , \sqrt{30} , \sqrt{38}}$