#sqrt(x^2+4x-21)+sqrt(x^2-x-6)=sqrt(6x^2-5x-39)#
#(sqrt(x^2+4x-21)+sqrt(x^2-x-6))^2=(sqrt(6x^2-5x-39))^2#
#x^2+4x-21+2*sqrt(x^2+4x-21)*sqrt(x^2-x-6)+x^2-x-6=6x^2-5x-39#
#2x^2+3x-27+2sqrt((x^2+4x-21)(x^2-x-6))=6x^2-5x-39#
#2sqrt((x^2+4x-21)(x^2-x-6))=4x^2-8x-12#
#sqrt((x^2+4x-21)(x^2-x-6))=2x^2-4x-6#
#(sqrt((x^2+4x-21)(x^2-x-6)))^2=(2x^2-4x-6)^2#
#(x^2+4x-21)(x^2-x-6)=(2)^2(x^2-2x-3)^2#
#(x+7)(x-3)(x-3)(x+2)=4((x-3)(x+1))^2#
#(x+7)(x-3)^2(x+2)=4(x-3)^2(x+1)^2#
#(x-3)^2((x+7)(x+2)-4(x+1)^2)=0#
#(x-3)^2(x^2+9x+14-4(x^2+2x+1))=0#
#(x-3)^2(x^2+9x+14-4x^2-8x-4)=0#
#(x-3)^2(-3x^2+x+10)=0#
#-(x-3)^2(3x^2-x-10)=0#
#-(x-3)^2(3x+5)(x-2)=0#
#x=3# or #x=-5/3# or #x=2#
However, as @Mark D points out, all the solutions but #x=3# give negative numbers within the square roots and so only #x=3# is valid:
#sqrt(x^2-x-6)#
#sqrt(2^2-2-6)=>sqrt(4-2-6)=>sqrt(-4)#
#sqrt((-5/3)^2+5/3-6)=>sqrt(25/9+15/9-54/9)=>sqrt(-14/9)#
