The equation √$(x+7)=x-5$ is satisfied by:

Squaring gives $x+7=x^2-10x+25$, i.e. $x^2-11x+18=0$, so $x=2$ or $x=9$. Test both in the original equation: for $x=2$, √9 = 3 but $x-5=-3$, so 2 is an extraneous root introduced by squaring; for $x=9$, √16 = 4 = 9 − 5. Only $x=9$ works.