Troubles with solving $\sqrt{2x+3}-\sqrt{x-10}=4$ I have been trying to solve the problem $\sqrt{2x+3}-\sqrt{x-10}=4$ and I have had tons problems of with it and have been unable to solve it. Here is what I have tried-$$\sqrt{2x+3}-\sqrt{x-10}=4$$ is the same as $$\sqrt{2x+3}=4+\sqrt{x-10}$$ from here I would square both sides $$(\sqrt{2x+3})^2=(4+\sqrt{x-10})^2$$
which simplifies to $$2x+3=16+x-10+8\sqrt{x-10}$$ I would then isolate the radical $$x-3=8\sqrt{x-10}$$ then square both sides once again $$(x-3)^2=(8\sqrt{x-10})^2$$ which simplifies to $$x^2-6x+9=8(x-10)$$ simplified again $$x^2-6x+9=8x-80$$ simplified once again $$x^2-14x+89=0$$ this is where I know I have done something wrong because the solution would be $$14 \pm\sqrt{-163 \over2}$$ I am really confused and any help would be appreciated

Let $~t=\sqrt{x-10}.~$ Then $~x=t^2+10.~$ Replacing, we have $~\sqrt{2t^2+23}-t=4,~$ which can be 
rewritten as $~\sqrt{2t^2+23}=t+4.~$ Squaring both sides, we have $~2t^2+23=t^2+8t+16.~$ Then, 
subtracting, we are left with $~t^2-8t+7=0,~$ whose two roots are $~t=1~$ and $~t=7,~$ where 
the former returns $~x=11,~$ while the latter yields $~x=59.~$ The reason why I wanted to share 
this method with you is because, in my opinion, it is less confusing $~($and therefore less prone to 
basic calculation errors$)~$ than the one you already tried.
