Verified Answer :

crisforp

2^(2x) = (2^x)^2 ;
2^(x+2) = (2^x) × 4 ;
Let be 2^x = y > 0 ;
y^2 + 4y - 12 = 0 ;
a = 1 ;
b = 4 ;
c = - 12 ;
b^2 - 4ac = 16 + 48 = 64 ;
y1 = ( - 4 + 8 ) / 2 = 2 > 0 ( correct ) ;
y2 = ( - 4 - 8 ) / 2 = - 6 > 0 ( false ) ;
Then, 2^x = 2 ;
x = 1.

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