Question 588770
Something is wrong with the problem, or my interpretation.
I cannot find a real x that is a solution of that equation.
If there is no x, how could there be an {{{x^2-3x+3}}} ?
If {{{ sqrt (x^2 -5x -11) + 13 =x }}} <--> {{{ sqrt (x^2 -5x -11)=x-13 }}} and we are working in the real numbers,
then {{{ x^2 -5x -11>=0}}} and {{{x>=13}}}.
Then, squaring both sides, we get
{{{x^2 -5x -11=(x-13)^2 }}} --> {{{x^2 -5x -11=x^2-26x+169}}} --> {{{-5x-11=-26x+169}}} --> {{{-5x-11+26x+11=-26x+169+26x+11}}} --> {{{21x=180}}} --> {{{x=180/21}}} --> {{{x=60/7}}}
But that is only a solution of {{{x^2 -5x -11=(x-13)^2 }}} 
It is not a solution of {{{ sqrt (x^2 -5x -11) + 13 =x }}} <--> {{{ sqrt (x^2 -5x -11)=x-13 }}}
because it makes {{{x-13<0}}}