Question 322184
{{{sqrt(x-5) = x-7}}} : x>7 (need positive), 

square : {{{x-5=x^2-14x+49}}} => {{{x^2-15x+54=0}}}

=>{{{x=(15 +- sqrt(15^2-216))/2=(15 +- 3)/2}}}=9 or 6

hence x=9 (to be >7)