The given equation is
$\sqrt{{\left(\sqrt{\mathrm{\text{x}}-2}-4\right)}^2}+\sqrt{{\left(\sqrt{\mathrm{\text{x}}-2}-5\right)}^2}=3$
$\Rightarrow \left|\sqrt{\mathrm{\text{x}}-2}-4\right|+\left|\sqrt{\mathrm{\text{x}}-2}-5\right|=3$
Put $\sqrt{\mathrm{\text{x}}-2}=\mathrm{\text{y}}$

Then, the given equation becomes $\left|\mathrm{\text{y}}-4\right|+\left|\mathrm{\text{y}}-5\right|=3$
Case $1:\mathrm{\text{y}}<4\Rightarrow -\text{y}+4-\text{y}+5=3$
$\Rightarrow -2\mathrm{\text{y}}=-6\Rightarrow \mathrm{\text{y}}=3$
Case $2: 4\le \text{y}<5\Rightarrow \text{y}-4-\text{y}+5=3$

$\Rightarrow 1=3\Rightarrow$ no solution
Case $3:$ $\text{y}\ge 5\Rightarrow \text{y}-4+\text{y}-5=3$
$\Rightarrow 2\mathrm{\text{y}}=12\Rightarrow \mathrm{\text{y}}=6$
$\mathrm{\text{y}}=3\Rightarrow \sqrt{\mathrm{\text{x}}-2}=3\Rightarrow \mathrm{\text{x}}-2=9\Rightarrow \mathrm{\text{x}}=11$
$\mathrm{\text{y}}=6\Rightarrow \sqrt{\mathrm{\text{x}}-2}=6\Rightarrow \mathrm{\text{x}}-2=36\Rightarrow \mathrm{\text{x}}=38$
$\Rightarrow \mathrm{\text{x}}=11$ and $38$ i.e. two solutions.