How do we solve $x^2 + \{x\}^2 = 33$ without computer? This is a problem taken from a group on Facebook. I wonder how to solve this without numerical process.
$x^2 + \{x\}^2 = 33\tag{1}$
My unfinished attempt:
$$\begin{align}
x^2 + \{x\}^2 &= 33\\
x^2 + \left(x - \lfloor x \rfloor\right)^2 &= 33\\
x^2 + x^2 - 2x \lfloor x \rfloor + \lfloor x \rfloor^2 &= 33\\
2x\left(x - \lfloor x \rfloor\right) + \lfloor x \rfloor^2 &= 33\\
2x\{x\} + \lfloor x \rfloor^2 = 33
\end{align}$$
I'm stuck at there. Also we know that the fractional part of $x$ is bounded i.e.:
$$0\leq\{x\}<1$$
From that, I can predict where the two solutions are placed at:
$$\begin{align}
0&\leq \{x\} < 1\\
0 &\leq \{x\}^2 < 1\\
x^2 &\leq x^2 + \{x\}^2 < x^2 + 1\\
x^2 &\leq 33 < x^2 + 1 \tag{$x^2 + \{x\}^2 = 33$}\\
S &= -\sqrt{33}\leq x \leq \sqrt{33} \quad \bigcap \quad  x< -\sqrt{32} \lor x > \sqrt{32}\\
\therefore S &= -\sqrt{33}\leq x < -4\sqrt2  \quad \bigcup \quad  4\sqrt{2} < x \leq \sqrt{33}\\
S &\approx -5.74456 \leq x < -5.65685  \quad \bigcup \quad  5.65685 < x \leq 5.74456
\end{align}$$
I don't know if that's even going to help. Anyway, here are the two solutions (to the original problem) that's given by Wolfram Alpha:
$$\begin{align}
x_1 &= \frac12 \left(3\sqrt{17} - 1\right) \approx -5.815\\
x_2 &= \frac12 \left(-1 - 3\sqrt{113}\right) \approx 5.685
\end{align}$$
You see $x_1$ isn't at the interval. I'm confused.

${x}^{\mathrm{2}} +\left\{{x}\right\}^{\mathrm{2}} =\mathrm{33} \\ $
$-\left\{{x}\right\}\pm\sqrt{\mathrm{33}−\left\{{x}\right\}^{\mathrm{2}} }=\left[{x}\right]\in\left\{-\mathrm{6},\mathrm{5}\right\} \\ $
$\mathrm{33}−{x}^{\mathrm{2}} \in\left\{\left({x}+\mathrm{6}\right)^{\mathrm{2}} ,\left({x}−\mathrm{5}\right)^{\mathrm{2}} \right\} \\ $
$\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{3}\right)\left({x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{4}\right)=\mathrm{0} \\ $
${x}\in\left\{-\mathrm{3}−\frac{\sqrt{\mathrm{30}}}{\mathrm{2}},\frac{\mathrm{5}+\sqrt{\mathrm{41}}}{\mathrm{2}}\right\} \\ $
