Floor it!

We know $\lfloor x \rfloor\leq x$

$\Rightarrow -8\lfloor x \rfloor\geq -8x$

$\Rightarrow x^{2}-8\lfloor x \rfloor+7\geq x^{2}-8x+7$

Let $f(x)=x^{2}-8\lfloor x \rfloor+7$

and $g(x)=x^{2}-8x+7$

$\Rightarrow f(x)\geq g(x)$

For $x\geq 8$

$\Rightarrow g(x)\geq 7$

$\Rightarrow f(x)\geq 7$ as $f(x)\geq g(x)$

But we want the value of $x$ such that $f(x)=0$ which is less than 7

$\Rightarrow x<8$

$\Rightarrow \lfloor x \rfloor \in {0,1,2,3,4,5,6,7}$

Since $x^{2}=8\lfloor x \rfloor-7$

Hence $x^{2} \in {-7,1,9,17,25,33,41,49}$

Since $x^{2}\geq0$ Therefore $x^{2} \in 1,9,17,25,33,41,49$

Checking against the equation leaves $x\in 1, \sqrt[]{33}, \sqrt[]{41}, 7$

So the answer is $\boxed{4}$