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?

From where you left off, squaring both sides again yields
$(2s + 2)^2 = (4 √(7-s))^2$ Which becomes
$4s^2 +8s +4 = 16(7-s)$
And you have a quadratic to solve, take all terms to one side and factorise or use the quadratic formula to find solutions to $s$, there will be two solutions of course. And then plug values of $s$ found back into your original equation to make sure you dont have a negative value under the square root.
