<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] $$2s+2=4\sqrt{7-s}$$
Divide by $2$ both sides
$$s+1=2\sqrt{7-s}$$
Square both sides
$$s^2+1+2s=28-4s$$
Put everything on the left
$$s^2+6s-27=0$$
Now solve by radicals 
$$s_{1,2}=\frac{-6\pm\sqrt{6^2-4(-27)}}{2}$$
giving you the two solutions
$$s_1=3\qquad;\qquad s_2=-9$$
Finally discard $s_2$ since it is not a solution of your title equation
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