Find sum of real solutions $16\times4^{2x}-(16x-47)4^x=96x^2+221x+63$ 
Find sum of real solutions of this equation: $16\times4^{2x}-(16x-47)4^x=96x^2+221x+63$

My attempt:
I rewrited the expression: $$16\times4^{2x}-(16x-47)4^x=(3x+1)(32x+63)$$
but its not really helping and by taking the function $f(x)=16\times4^{2x}-(16x-47)4^x-(3x+1)(32x+63)$ and trying to study it and taking the derivative it's not really pleasant. Guessed a solution $x=0$.
 A: take $w=4^x$  so that the quadratic formula for $16 w^2 -(16x-47)w - (96x^2+221x+63)=0 \; \; . $ 
The part under the square root becomes
$$ \sqrt{ ( 16  x - 47)^2 + 64 (96  x^2 + 221  x + 63   ) \; \;} = \sqrt{6400x^2 + 12640x + 6241 \; \; }  $$
which
gives $80x+79,$ so 
$$ w = \frac{16x-47 \pm (80x+79)}{32}   $$
With the plus sign, we get
$$ 4^x = 3 x + 1  $$
so $$  x = 0,1  $$
With the minus sign, we get
$$ 4^x = -2x - 4 + \frac{1}{16} $$
so $$  x=-2 $$
then
$$ 0+1-2 = -1 $$
A: Calling $y = 4^x$ we have
$$
16 y^2 - (16 x - 47) y -( 96 x^2 + 221 x + 63) = 16\left(y-1-3x\right)\left(y+\frac{63+32x}{16}\right)=0
$$
or
$$
4^x-1-3x = 0\\
4^x+\frac{63+32x}{16} = 0
$$
Those equations can be solved with the contribution of the Lambert function giving
$$
x \in \{-2,0,1\}
$$
A: Thanks @Cesare 
its easy to see the answer is $-79/80$, by comparing the two equations
$4^x-1-3*x = 0$.
$x+\frac{63+32*x}{16} = 0$
one can get one of the real value of $x$by putting $4^x = 1+3*x$ in the second equation. and in turn to get more solutions easily we can say $x = \frac{4^x-1}{3}$.
then $x = \log_(4) (141/80)$
