You complete the square with a negative $x^2$ in place.
Factor the $-1$ out first: $-x^2+4x+5 = -(x^2-4x)+5 = -(x-2)^2+9$. The resulting sign decides the standard form — $a^2-u^2$ leads to an inverse sine, $a^2+u^2$ to an inverse tangent.
Next time: Negative coefficient on $x^2$ means pull the $-1$ out first.