Looks like the labeled angle is bisected, in which case we can apply the angle bisector theorem. For this triangle, it says
[tex]\dfrac x3=\dfrac{x+2}{x+4}[/tex]
Solve for [tex]x[/tex]:
[tex]x(x+4)=3(x+2)[/tex]
[tex]x^2+4x=3x+6[/tex]
[tex]x^2+x-6=0[/tex]
[tex](x+3)(x-2)=0[/tex]
Then either [tex]x=-3[/tex] or [tex]x=2[/tex], but since [tex]x[/tex] is a length, it must be positive. So x = 2.
