How to solve $2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$ for $x$? I'm struggling to figure out how to solve $2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$ for $x$. 
The answer is $x = \frac{12}{5}$, but I am getting $x = \frac{-35}{9}$
My steps are:
$2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$


*

*make exponent postive


$2x\frac{1}{\sqrt{4-x}} - 3\sqrt{4-x} = 0$


*

*square everything 


$4x^2\frac{1}{4-x} - 9(4-x) =0$


*

*combine into one fraction


$\frac{1-9(4-x)[4x^2(4-x)]}{4x^2(4-x)} = 0$


*

*remove like terms


$1-9(4-x)=0$
$1-36-9x = 0$
$x = \frac{-35}{9}$

If we put $x=4t$, the equation becomes
$$\frac{4t}{\sqrt{1-t}}=6\sqrt{1-t}$$
and
$$2t=3(1-t)$$
which gives
$$t=\frac 35 \; ,\; x=\frac{12}{5}.$$
