Question 1019621
Your text is not well written so one could guess you have sqrt(2x)+x=4, which would best appear {{{sqrt(2x)+x=4}}}.


Isolate the radical.
{{{sqrt(2x)=4-x}}}


Square the equated members.
{{{2x=(4-x)^2}}}
{{{2x=16-8x+x^2}}}
Simplify and solve.
{{{x^2-8x-2x+16=0}}}
{{{x^2-10x+16=0}}}
{{{(x-2)(x-8)=0}}}
and if you check you find both x=2  and x=8 will work.
x=2  OR  x=8.