Quadratic Equation?

Note that $\displaystyle \left(2\sqrt{\frac{x}{8}+\frac{13}{64}}-\frac{1}{4}\right)^2=4\left(\frac{x}{8}+\frac{13}{64}\right)-\sqrt{\frac{x}{8}+\frac{13}{64}}+\frac{1}{16}=\frac{x}{2}+\frac{7}{8}-\sqrt{\frac{x}{8}+\frac{13}{64}}$ .

It is clear $x>0$ and $\displaystyle 2\sqrt{\frac{x}{8}+\frac{13}{64}}>2\sqrt{\frac{13}{64}}>\frac{1}{4}$ and hence $\displaystyle \sqrt{\frac{x}{2}+\frac{7}{8}-\sqrt{\frac{x}{8}+\frac{13}{64}}}=2\sqrt{\frac{x}{8}+\frac{13}{64}}-\frac{1}{4}$ .

Now $\displaystyle x-\sqrt{\frac{x}{2}+\frac{7}{8}-\sqrt{\frac{x}{8}+\frac{13}{64}}}=x-\left(2\sqrt{\frac{x}{8}+\frac{13}{64}}-\frac{1}{4}\right)=x+\frac{1}{4}-\sqrt{\frac{x}{2}+\frac{13}{16}}$ .

Note that $\displaystyle \left(\sqrt{2}\sqrt{\frac{x}{2}+\frac{13}{16}}-\frac{1}{2\sqrt{2}}\right)^2=\left(x+\frac{13}{8}\right)-\sqrt{\frac{x}{2}+\frac{13}{16}}+\frac{1}{8}=\left(x+\frac{1}{4}-\sqrt{\frac{x}{2}+\frac{13}{16}}\right)+\frac{3}{2}$ .

Now $\displaystyle x-\sqrt{\frac{x}{2}+\frac{7}{8}-\sqrt{\frac{x}{8}+\frac{13}{64}}}=\left(\sqrt{2}\sqrt{\frac{x}{2}+\frac{13}{16}}-\frac{1}{2\sqrt{2}}\right)^2-\frac{3}{2}=179$

Hence, $\displaystyle \left(\sqrt{x+\frac{13}{8}}-\frac{1}{2\sqrt{2}}\right)^2=179+\frac{3}{2}=\frac{361}{2}=\left(\frac{19}{\sqrt{2}}\right)^2$

Finally, solving it to obtain $\displaystyle x=\left(\frac{19}{\sqrt{2}}+\frac{1}{2\sqrt{2}}\right)^2 -\frac{13}{8}=188.5$ and hence $\displaystyle 10x=1885$ .

Let $u=\sqrt{\dfrac{x}{8}+\dfrac{13}{64}}$

$\implies 8u^2-\dfrac{13}{8}-\sqrt{4u^2-\dfrac{13}{16}+\dfrac{7}{8}-u}=179$

$8u^2-\dfrac{13}{8}-\sqrt{4u^2-u+\dfrac{1}{16}}=179$

$8u^2-\dfrac{13}{8}-\left|2u-\dfrac{1}{4}\right|=179$

Since $x\geq 179\implies$ $u\geq \dfrac{17\sqrt{5}}{8}$ and $2u>\dfrac{1}{4}$

$\implies 8u^2-\dfrac{13}{8}-\left(2u-\dfrac{1}{4}\right)=179$

$8u^2-2u-\dfrac{1443}{8}=0$

$\implies u=\dfrac{39}{8}=\sqrt{\dfrac{x}{8}+\dfrac{13}{64}}$

$\implies x=\dfrac{377}{2}$

$\therefore \boxed{10x=1885}$