jmc

Completing the square in $x$ and $y,$ we get $$2(x-2{)}^2+3(y+1{)}^2=6.$$Then $$\frac{(x-2{)}^2}{3}+\frac{(y+1{)}^2}{2}=1.$$Thus, the center of the ellipse is $(2,-1),$ and the semi-axis in the $x$-direction is $\sqrt{3}.$



Hence, the maximum $x$-coordinate of a point on the ellipse is $\boxed{2+\sqrt{3}}.$