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

Note that when you square something like $a\sqrt{b}$ you get $a^2b$.
Thus, you should get:
$\begin{align}
x^2-6x+9 &= 64(x-10)\\
x^2-6x+9 &= 64x-640\\
x^2-70x+649 &= 0\\
(x-11)(x-59) &= 0\\
\therefore \boxed{x=11,59}.
\end{align}$
