$y=\{x\}$ represents greatest integer less than $x$ then solve $x^2-7\{x\}+5=0$. $y=\{x\}$ represents greatest integer less than $x$,
e.g. 


*

*$\{1.1\}=1$

*$\{2\}=1$

*$\{5\}=4$

*$\{3.7\}=3$


Then what are the  solutions of
$$x^2-7\{x\}+5=0?$$
Note that this is not same as greatest integer function. So its more difficult to solve.

Your problem is equivalent to solving
$$ x^2-7\lceil x\rceil +12=0. $$
If $x\le 1$, then $$x^2-7\lceil x\rceil+12\ge -7\lceil x\rceil +12\ge5$$
hence we obtain $x>1$.
If $x>7$, then using $\lceil x\rceil < x+1 $ we get $$x^2-7\lceil x\rceil +12>x^2-7x+5=(x-7)x+5>0.$$
We conclude $\lceil x\rceil \in\{2,3,4,5,6,7\}$ and accordingly $x^2=7\lceil x\rceil -12\in\{2,9,16,23,30,37\}$. As $x>0$, we conlcude $$x\in\{\sqrt 2, 3, 4, \sqrt{23},\sqrt{30}\sqrt{37}\}.$$ Checking against the original equation shows that all of these are in fact valid solutions 
