Answer
$(2,\sqrt 2)$
Work Step by Step
$y= \sqrt x$ Equation $(1)$
$x^{2}+y^{2}= 6$ Equation $(2)$
Substitute $y= \sqrt x$ in Equation $(2)$ we get,
$x^{2}+y^{2}= 6$
$x^{2}+(\sqrt x)^{2}= 6$
$x^{2}+x= 6$
$x^{2}+x- 6=0$
By factoring,
$(x-2)(x+3)=0$
$x=2$ or $x=-3$
Substitute $ x$ values in Equation $(1)$ to get $y$,
Let $x=2$
$y= \sqrt 2$
Let $x=-3$
$y= \sqrt (-3)$ is not a real number.
$(2,\sqrt 2)$ satisfies both equations. So, $(2,\sqrt 2)$ is the only solution.