Question 837035
{{{sqrt(2x-7)^2=(5-x)^2}}}
{{{2x-7=25-10x+x^2}}}
{{{x^2-12x+32=0}}}
{{{(x-4)(x-8)=0}}}


but if x=8 then 5-8=-3 ... impossible, so...

solution:
x=4