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.