First of all, you should know that any number of the form a^(bc) can be written in the form (a^b)^c or (a^c)^b or vice-versa.
i.e. a^(bc) = (a^b)^c = (a^c)^b.
Also, a^(b+c) = a^b*a^c
therefore above equation 2^(2x)-(2^(x+2))-32=0 can be writte as:-
(2^x)^2 - 2^2*2^x-32=0
=> (2^x)^2-4*2^x-32=0
Let 2^x be some number y.
therefore y^2-4y-32=0
Now solving it just like a regular quadratic equation
=> y^2-8y+4y-32=0
=> y(y-8)+4(y-8)=0
=> (y+4)(y-8)=0
y=-4, 8
But, y = 2^x
=> 2^x = -4, 8
Since any exponential function defined on real numbers cannot have a negative value, we will discard -4.
graph{2^x [-10, 10, -5, 5]}
You can observe in this graph of 2^x that no matter what the value of x, the value of y=2^x is always positive. Its value is getting closer and closer to zero but is never actually touching zero. This is true for any function of the type a^x where a and x are REAL Numbers.
therefore 2^x = 8
=> 2^x = 2^3
=> x=3