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
 A: Note that when you square something like $a\sqrt{b}$ you get $a^2b$.
Thus, you should get:
$\begin{align}
x^2-6x+9 &= 64(x-10)\\
x^2-6x+9 &= 64x-640\\
x^2-70x+649 &= 0\\
(x-11)(x-59) &= 0\\
\therefore \boxed{x=11,59}.
\end{align}$
A: 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.
A: Alternatively, take the reciprocal to get:
$$\frac{1}{\sqrt{2x + 3} - \sqrt{x - 10}} = \frac{\sqrt{2x + 3} + \sqrt{x - 10}}{(2x+3)-(x-10)} = \frac{\sqrt{2x + 3} + \sqrt{x - 10}}{x+13}$$
which we know is equal to $\frac 1 4$.
Thus we have the following:
$$\sqrt{2x+3} - \sqrt{x+10} = 4 \tag{1}$$
$$\sqrt{2x + 3} + \sqrt{x - 10} = \frac{1}{4}(x+13) \tag{2}$$
Add the two together to get:
$$2 \sqrt{2x + 3} = 4 + \frac{1}{4}(x + 13)$$
$$8\sqrt{2x+3} = 16 + (x + 13)$$
$$64(2x + 3) = x^2 + 58x + 841$$
$$x^2-70x+649=0$$
$$(x - 11)(x - 59) = 0 \implies x = \boxed{11, 59}.$$
