My solution:
First, notice that if we rewrite the equality as \displaystyle 4x^2+51=40\left\lfloor{x}\right\rfloor, we can tell \displaystyle \left\lfloor{x}\right\rfloor must be a positive figure.
Second, we let \displaystyle x=\left\lfloor{x}\right\rfloor+\{x\}, where \{x\} represents the decimal part of x.
This transforms the original equation to become:
\displaystyle 4x^2-40\left\lfloor{x}\right\rfloor+51=0
\displaystyle 4(\left\lfloor{x}\right\rfloor+\{x\})^2-40\left\lfloor{x}\right\rfloor+51=0
\displaystyle 4(\left\lfloor{x}\right\rfloor^2+2\left\lfloor{x}\right\rfloor\{x\}+\{x\}^2)-40\left\lfloor{x}\right\rfloor+51=0
\displaystyle (4\left\lfloor{x}\right\rfloor^2-40\left\lfloor{x}\right\rfloor+51)+8\left\lfloor{x}\right\rfloor\{x\}+2\{x\}^2=0
\displaystyle (2\left\lfloor{x}\right\rfloor-17)(2\left\lfloor{x}\right\rfloor-3)+2\{x\}(4\left\lfloor{x}\right\rfloor+\{x\})=0
Note that since \displaystyle 2\{x\}(4\left\lfloor{x}\right\rfloor+\{x\})\ge 0, we know \displaystyle (2\left\lfloor{x}\right\rfloor-17)(2\left\lfloor{x}\right\rfloor-3)+2\{x\}(4\left\lfloor{x}\right\rfloor+\{x\}) must be a negative.
Thus, solving \displaystyle (2\left\lfloor{x}\right\rfloor-17)(2\left\lfloor{x}\right\rfloor-3)+2\{x\}(4\left\lfloor{x}\right\rfloor+\{x\})\le 0 for \displaystyle \left\lfloor{x}\right\rfloor, we see that we have:
\displaystyle 2\le \left\lfloor{x}\right\rfloor \le 8.
Checking for each case of \displaystyle \left\lfloor{x}\right\rfloor=2,\,3,\,4,\,5,\,6,\,7,\,8, we find only \displaystyle \left\lfloor{x}\right\rfloor=2,\,6,\,7,\,8 work, which yields
\displaystyle x=\frac{\sqrt{29}}{2},\,\frac{3\sqrt{21}}{2},\,\frac{\sqrt{229}}{2},\,\frac{\sqrt{269}}{2}.
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