Taking squares on both sides leads to
[tex]\sqrt{-x + 6} = x - 6[/tex]
[tex]\left(\sqrt{-x + 6}\right)^2 = (x - 6)^2[/tex]
[tex]-x + 6 = x^2 - 12x + 36[/tex]
[tex]x^2 - 11x + 30 = 0[/tex]
[tex](x - 5) (x - 6) = 0[/tex]
Solving for [tex]x[/tex], we get
[tex]x - 5 = 0 \text{ or } x - 6 = 0[/tex]
or
[tex]x = 5 \text{ or } x = 6[/tex].
Evaluating both sides of the starting equation at these solutions, we have
[tex]\sqrt{-6 + 6} = 6 - 6 \implies 0 = 0[/tex]
which is true, so [tex]\boxed{x=6}[/tex] is a valid solution. However,
[tex]\sqrt{-5 + 6} = 5 - 6 \implies \sqrt1 = -1 \implies 1 = -1[/tex]
which is not true, so [tex]\boxed{x=5}[/tex] is an extraneous solution.
