#sqrt(x-3)=sqrt(4(x+4))-1#
Since we cannot get rid of the #1#, let's square and see what we get.
#(sqrt(x-3))^2=(sqrt(4(x+4))-1)^2#
#x-3=(sqrt(4(x+4)))^2-2*sqrt(4(x+4))*1+(1)^2#
#x-3=4(x+4)-2sqrt(4(x+4))+1#
#x-3=4x+17-2sqrt(4(x+4))#
#2sqrt(4(x+4))=3x+20#
Now, let's square again.
#(2sqrt(4(x+4)))^2=(3x+20)^2#
#4*4(x+4)=(3x)^2+2*(3x)*20+20^2#
#16x+64=9x^2+120x+400#
#9x^2+104x+336=0#
Before we use the quadratic formula, let's check the discriminant to see if there are real solutions. If #b^2-4ac# is less than #0#, then there are no real solutions; if #0#, there is only #1# solution; if greater than #0#, there are #2# solutions. You can analyze the quadratic formula to see why this is the case.
#(104)^2-4*9*336#
#=-1280#
Thus, there are no real solutions.
