Solve the quadratic with GIF

We know $[x]\leq x$

$\Rightarrow -7[x]\geq -7x$

$\Rightarrow x^{2}-7[x]+5\geq x^{2}-7x+5$

Let $f(x)=x^{2}-7[x]+5$

and $g(x)=x^{2}-7x+5$

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

For $x\geq 7$

$\Rightarrow g(x)\geq 5$

$\Rightarrow f(x)\geq 5$ as $f(x)\geq g(x)$

But we want the value of $x$ such that $f(x)=0$ which is less than 5

$\Rightarrow x<7$

$\Rightarrow [x] \in {0,1,2,3,4,5,6}$

Since $x^{2}=7[x]-5$

Hence $x^{2} \in {-5,2,9,16,23,30,37}$

Since $x^{2}\geq0$ Therefore $x^{2} \in 2,9,16,23,30,37$

So now we have $x\in \sqrt[ ]{2},3,4, \sqrt[ ]{23}, \sqrt[ ]{30},\ \sqrt[ ]{37}$

Checking against the equation leaves $x\in \sqrt[]{2}, \sqrt[]{23}, \sqrt[]{30}, \sqrt[]{37}$

Therefore the sum of squares is $2+23+30+37=\boxed{92}$

We can express $x$ in terms of its integral part $\lfloor x \rfloor$ and up from there draw some conclusions:

$x^2-7\lfloor x \rfloor +5=0$

$x= \pm \sqrt[]{7\lfloor x \rfloor-5}$

$\left \lfloor \pm \sqrt[]{7\lfloor x \rfloor-5} \right \rfloor = \lfloor x \rfloor \; \Longrightarrow \; \lfloor x \rfloor \le \pm \sqrt[]{7\lfloor x \rfloor-5} < \lfloor x \rfloor +1$

When then solving the inequality for $\lfloor x \rfloor$ , if we choose the negative sign we see that there are no solutions. As for $+$ instead, we get $\frac{7-\sqrt{29}}{2} \le \lfloor x \rfloor <2 \, \lor \, 3 < \lfloor x \rfloor \le \frac{7+\sqrt{29}}{2}$ and so rounding: $0.81 \le \lfloor x \rfloor < 2 \, \lor \, 3< \lfloor x \rfloor \le 6.19$ .

Hence, $\lfloor x \rfloor =$ $1$ , $4$ , $5$ , $6 \,$ and the value of the sum is

$\displaystyle \sum_{x \, \text{is a root}} x^2 = (7 \cdot 1-5)+( 7 \cdot 4-5 ) + (7 \cdot 5-5 )+( 7 \cdot 6-5 )= \boxed{92}$