Question 989398
This solution requires squaring twice.

{{{sqrt(2x-3)=sqrt(x-2)+1}}}
{{{2x-3=x-2+2sqrt(x-2)+1}}}
.
.
{{{x-2=2sqrt(x-2)}}}
{{{x^2-4x+4=4(x-2)}}}
{{{x^2-8x+12=0}}}


{{{highlight_green((x-2)(x-6)=0)}}}


Either {{{x=2}}}  OR  {{{x=6}}}.
Either of them will work.