Question 442311
If I understand correctly your two problems are:

1)  {{{sqrt(2x) = -4}}} 

2) {{{sqrt(x+222) - sqrt(x+46) = 8}}}

For number 1, remember that to "undo" square roots, we need to square both sides.
Thus,  we have {{{2x = 16}} so x =8. But if we think about it, there is no way using real numbers that we can obtain a negative number by using the square root. If we were to use imaginary numbers where  {{{i  = sqrt(-1)}}}, then our answer would be {{{x = 8i}}}.

For number 2, let's multiply by the conjugate.

{{{(sqrt(x+222)-sqrt(x+46)) * (sqrt(x+222) + sqrt(x+46))/(sqrt(x+222) + sqrt(x+46))}}}

{{{x+222 - x+46  = 8(sqrt(x+222) + sqrt(x+46))}}}

{{{176 = 8(sqrt(x+222) + sqrt(x+46))}}}

{{{22 = sqrt(x+222) + sqrt(x+46))}}}

recall that {{{ 8=sqrt(x+222) - sqrt(x+46)}}}

Then    {{{22 = sqrt(x+222) + sqrt(x+46)}}}
         {{{8 = sqrt(x+222) - sqrt(x+46)}}}

if we add the two equations we get:

30 = 2sqrt(x+222)
15 = sqrt(x+222)
225 = x+222
x =3.

if we subtract the two equations we get:

14 = 2sqrt(x+46)
7 = sqrt(x+46)
49 = x + 46
x=3

So, {{{x = 3}}}