<s>[INST] Solve equation $\sqrt{s+13} - \sqrt{7-s} = 2$ Solve the equation 
$$\sqrt{s+13}-\sqrt{7-s} = 2$$
I moved the $-\sqrt{7-s}$ to the right side 
Thus, I had 
$$\sqrt{s+ 13} = 2 +\sqrt{7-s}$$
I then squared both sides
$$\sqrt{s+ 13}^2 = \left(2 +\sqrt{7-s}\right)^2$$
Using the product formula $(x + y)^2 = x^2 + 2xy + y^2$
I got  $$s + 13 = 4 + 4\sqrt{7-s}+  7 – s$$
I then combined like terms 
$$2s + 2= 4 \sqrt{7-s}$$
I’m stuck at this point. Does anyone have an idea how to solve this equation?
 [/INST] You're almost at the end result:
After achieving $ 2s + 2 = 4 \sqrt{7-s} $, square both sides again to generate a quadratic equation.
$$ 4 ( s + 1 )^2 = 16 ( 7 - s ) = 4 (s^2 + 2s+ 1)$$
Solving which is quite easy.
$$\begin{align}
s^2 + 2s + 1 &= 28 - 4s \\
s^2 + 6s - 27 &= 0 \\
(s + 9) (s - 3) &= 0
\end{align}$$

NOTE
You have to neglect $ s = -9 $ from the solutions as it'll give you $ \sqrt{-9 + 13} - \sqrt{ 7 - (-9)} = \sqrt{16} = +2 - (+4) \neq 2 $
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