Case 1 : When $$\sqrt x \ge 2$$

then $$\left| {\sqrt x - 2} \right| = \sqrt x - 2$$

$$ \therefore $$ The given equation becomes,

$$\left( {\sqrt x - 2} \right)$$ + $$\sqrt x \left( {\sqrt x - 4} \right) + 2$$ = 0

$$ \Rightarrow $$ $$\left( {\sqrt x - 2} \right)$$ + $$x - 4\sqrt x $$ + 2 = 0

$$ \Rightarrow $$ $$x - 3\sqrt x $$ = 0

$$ \Rightarrow $$ $$\sqrt x \left( {\sqrt x - 3} \right)$$ = 0

$$ \therefore $$ $$\sqrt x $$ = 0 or 3

$$\sqrt x $$ = 0 is not possible as $$\sqrt x \ge 2$$.

So, $$\sqrt x $$ = 3

or $$x$$ = 9

Case 2 : When $$\sqrt x < 2$$

then $$\left| {\sqrt x - 2} \right| = $$$$ - \left( {\sqrt x - 2} \right)$$ = $$2 - \sqrt x $$

$$ \therefore $$ The given equation becomes,

$$\left( {2 - \sqrt x } \right)$$ + $$\sqrt x \left( {\sqrt x - 4} \right) + 2$$ = 0

$$ \Rightarrow $$ $${2 - \sqrt x }$$ + $$x - 4\sqrt x $$ + 2 = 0

$$ \Rightarrow $$ $$x - 5\sqrt x + 4$$ = 0

$$ \Rightarrow $$ $$x - 4\sqrt x - \sqrt x + 4$$ = 0

$$ \Rightarrow $$ $$\sqrt x \left( {\sqrt x - 4} \right)$$$$-\left( {\sqrt x - 4} \right)$$ = 0

$$ \Rightarrow $$ $$\left( {\sqrt x - 4} \right)$$$$\left( {\sqrt x - 1} \right)$$ = 0

$$ \therefore $$ $$\sqrt x $$ = 4 or 1

$$\sqrt x $$ = 4 is not possible as $$\sqrt x < 2$$.

$$ \therefore $$ $$\sqrt x $$ = 1

or $$x$$ = 1

So, Sum of all solutions = 9 + 1 = 10