[tex]2x(4-x)^{-1/2}-3(4-x)^{1/2}=(4-x)^{-1/2}\bigg(2x-3(4-x)\bigg)[/tex]
We effectively rewrite the equation as
[tex]\dfrac{5x-12}{(4-x)^{1/2}}=0[/tex]
In order for the LHS to be defined, we need to restrict [tex]4-x>0[/tex], or [tex]x<4[/tex]. Now, the LHS will vanish when the numerator is 0, which happens for
[tex]5x-12=0\implies5x=12\implies x=\dfrac{12}5[/tex]
This value is indeed smaller than 4, so the solution is [tex]x=\dfrac{12}5[/tex].
