Begin by finding the restrictions on the domain
#5+x>=0 and 3-x>=0#
Simplify
#x>=-5 and -x>=-3#
#x>=-5 and x<=3#
#-5<=x<=3#
Append the restrictions to the original equation:
#sqrt(5+x)-sqrt(3-x)=2;-5<=x<=3#
Square both sides:
#(sqrt(5+x)-(sqrt(3-x))^2=2^2;-5<=x<=3#
Expanding the square of the left side may appear to be a bit daunting but the pattern, #(a-b)^2 = a^2-2ab+b^2#, makes it quite simple:
Let #a = sqrt(5+x) and b = sqrt(3-x)#, then #a^2=5+x,b^2=3-x, and -2ab=-2sqrt((3-x)(5+x))#
#5+x -2sqrt((3-x)(5+x)) + 3-x=4;-5<=x<=3#
Collect everything but the radical term to one side:
#-2sqrt((3-x)(5+x)) = -4;-5<=x<=3#
Divide both sides by -2:
#sqrt((3-x)(5+x)) = 2;-5<=x<=3#
Square both sides:
#(3-x)(5+x)=4;-5<=x<=3#
Expand the left side, using the F.O.I.L. method:
#15+3x-5x-x^2=4;-5<=x<=3#
Combine like terms
#11-2x-x^2=0;-5<=x<=3#
Multiply both side by -1:
#x^2+2x-11=0;-5<=x<=3#
Check the discriminant:
#d=b^2-4(a)(c) = 2^2-4(1)(-11)=4+44=48#
Using the quadratic formula:
#x = (-b+-sqrt(d))/(2a)#
#x = (-2+-sqrt(48))/2#
#x = -1+-sqrt(12)#
#x = -1+-2sqrt(3)#
If you substitute the negative root into the original equation you obtain, #-2=2#, which indicates an extraneous root caused by squaring; therefore, it must be discarded:
#x = 2sqrt(3)-1# is the only solution.
