How do you solve #sqrt(x+4)-2=sqrt(x-12)#?

Question
Answer 1

#x=21#

#sqrt(x+4)-2 = sqrt(x-12)#
#(x+4)-4sqrt(x+4)+4=(x-12)#
Subtract #x# from both sides and simplify to get;
#8-4sqrt(x+4) = -12#
Subtract #8# from both sides to get:
#-4sqrt(x+4) = -20#
DIvide both sides by #-4# to get:
#sqrt(x+4) = 5#
#x+4 = 25#
Subtract #4# from both sides to get:
#x=21#
#sqrt(color(blue)(21)+4) - 2 = sqrt(25)-2 = 5-2 = 3 = sqrt(9) = sqrt(color(blue)(21)-12)#
So #x=21# is a solution of the original equation.
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