Question 1031834
{{{sqrt(4-2x) =  x+1}}}
square both sides
{{{4-2x =  (x+1)^2}}}
{{{4-2x =  x^2+ 2x + 1}}}
add 2x to each side
{{{4 =  x^2 + 4x + 1}}}
add -4 to each side
{{{0 =  x^2 + 4x - 3}}}
Using the quadratic formula we get
two roots, {{{-2 + sqrt(7)}}}
and {{{-2 -sqrt(7) }}}
Substituting {{{-2 -sqrt(7) }}} for x in the original equation does not work,
so our root is {{{-2 + sqrt(7)}}}