Finding $A$, $B$, $C$ such that $(2x-A)(2x+4)-Bx+25=(x-C)^2-Cx^2$ for all $x$ 
Find $A$, $B$, $C$ such that
$$(2x-A)(2x+4)-Bx+25=(x-C)^2-Cx^2$$
for all $x$.

This is what I've got so far.
$$\begin{align}
\text{L.H.S.} 
&= (2x-A)(2x+4)-Bx+25 \tag1\\
&= 4x^2-2Ax+8x-4A-Bx+25 \tag2\\
&= 4x^2-(2A+8)x-4A-Bx+25 \tag3 \\ \\
\text{R.H.S.}
&= (x-C)^2-Cx^2 \tag4\\
&= x^2-2Cx+C^2+Cx^2 \tag5
\end{align}$$
I have no idea how I should find the constants by calculation. Please help.

Since it is true for all $x$, this is an identity and hence all you need to do is compare coefficients as mentioned in comments. So, we have $$\begin{aligned} 4 &=1-C \\ 2A+B-8&=2C \\ -4A+25&=C^2\end{aligned}$$
Which gives you $C=-3,A=4,B=-6$.

Note that you have done some mistakes in sign.
