$\ \sqrt{x+39}-\sqrt{x+7}=4 $ So I tried to solve this problem for x
$\ \sqrt{x+39}-\sqrt{x+7}=4 $
I multiplied both sides ($\ \sqrt{m}\cdot\sqrt{n}=\sqrt{mn} $)
$\ (\sqrt{x+39}-\sqrt{x+7})^2=16 $
$\ (x+39)-2(x^2+46x+273)-(x+7) $
$\ 0x+32+(-2x^2-92x-546) $
$\ -2x^2-92x-514 $
divide the 2 out
$\ x^2+46x+257=-8 $
$\ x^2+46x+265=0 $
Use the quadratic formula (or scientific calculator) and the answers are -6.752 and -39.248. I know the answer is exactly -3. What went wrong?

One can square twice. I prefer to invert and obtain
$$\frac{1}{\sqrt{x+39}-\sqrt{x+7}}=\frac{1}{4},$$
and then by rationalizing the denominator get
$$\frac{\sqrt{x+39}+\sqrt{x+7}}{32}=\frac{1}{4},$$
or equivalently 
$$\sqrt{x+39}+\sqrt{x+7}=8.$$
Now "add" to the original equation, divide by $2$. We get $\sqrt{x+39}=6$, and now it's over. 
Remark: In the original post, the first step used was to square. This to some degree complicates things, the middle term should have been $-2\sqrt{(x+39)(x+7)}$. Rearrangement and squaring now get us to where we want.  
