Question 85463
Isolate the radicals by subtracting one of them from both sides:
{{{(sqrt(x+6))}}} = 4 - {{{(sqrt(2-x))}}} 
Square both sides:
{{{(sqrt(x+6))^2}}} = (4 - {{{(sqrt(2-x))}}})^2
x + 6 = 16 - 8{{{sqrt(2-x))}}} + 2-x
x + 6 = 18 - 8{{{sqrt(2-x))}}}-x
Isolate the remaining radical:
x + x + 6 - 18 = - 8{{{sqrt(2-x))}}}
2x-12 = - 8{{{sqrt(2-x))}}}
Square both sides again:
{{{(2x-12)^2}}} = (- 8{{{(sqrt(2-x))}}})^2
{{{4x^2}}} - 48x + 144 = 64(2 - x)
{{{4x^2}}}-48x+144 = 128-64x
{{{4x^2}}}+16x+16 = 0
{{{x^2}}} + 4x + 4 = 0
(x + 2)(x + 2) = 0
x + 2 = 0
x = -2 [potentially]
Check for extraneous solutions:
{{{(sqrt(-2+6))}}} = 4 - {{{(sqrt(2-(-2)))}}}?
{{{(sqrt(4))}}} = 4 - {{{(sqrt(4))}}} ?
2 = 4 - 2 ?
2 = 2 ; OK
x = -2 is a valid solution