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?
 A: $$2s+2=4\sqrt{7-s}$$
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
A: 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.
A: 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 $
A: Wolframalpha helps you solving equations and can even show you
a step-by-step solution.
A: Clearly $7\ge s\ge-13\iff7-(-3)\ge s+3\ge-13-(-3)$
WLOG $s+3=10\cos 2y$ where $0\le2y\le\pi$
$\implies\sqrt{s+13}=\sqrt{10(1+\cos2y)}=2\sqrt5\cos y,\sqrt{7-s}=2\sqrt5\sin y$
$\implies2\sqrt5\cos y-2\sqrt5\sin y=2$
Squaring we get $$20(1-\sin2y)=4\iff\sin2y=\dfrac45\implies\cos2y=\pm\sqrt{1-\sin^22y}=\cdots$$
Check which values of $s=10\cos2y-3$ satisfies the given equation.
