As currently written, solving for [tex]x[/tex], we have
[tex]\sqrt{15} - x + \sqrt3 - x = 6[/tex]
[tex]2x = \sqrt{15} + \sqrt3 - 6[/tex]
[tex]x = \dfrac{\sqrt{15} +\sqrt3 - 6}2[/tex]
I suspect you meant to write
[tex]\sqrt{15 - x} + \sqrt{3 - x} = 6[/tex]
Move one of the roots to the other side, then take squares on both sides.
[tex]\sqrt{15 - x} = 6 - \sqrt{3 - x}[/tex]
[tex]\left(\sqrt{15 - x}\right)^2 = \left(6 - \sqrt{3 - x}\right)^2[/tex]
[tex]15 - x = 36 - 12 \sqrt{3 - x} + (3 - x)[/tex]
[tex]12 \sqrt{3 - x} = 24[/tex]
Take squares again
[tex]\left(12\sqrt{3-x}\right)^2 = 24^2[/tex]
[tex]144 (3 - x) = 576[/tex]
[tex]432 - 144x = 576[/tex]
[tex]144x = -144[/tex]
[tex]\boxed{x = -1}[/tex]
