I don't want Integers

We note that $\space x^2-4x+5 = 2^{x-2} \quad \Rightarrow (x-2)^2+1 = 2^{x-2}$

Let $\space y=x-2\quad \Rightarrow f(y) = y^2 +1 -2^y = 0$

$\quad \Rightarrow f'(y)\frac {df(y)}{dy} = 2y - \ln{2}\dot{}2^y$

We note that: $f(4) = 1\space$ and $\space f(5) = -6\space$ implying there is a root within $4 < y < 5$ . Using Newton's method, we have:

$y_{n+1} = y_n - \dfrac {f_n(y)}{f_n'(y)}$

$\begin {matrix} y_n & f(y_n) & f'(y_n) & \frac {f(y_n)}{f'(y_n} \\ 4&1 & -3.090354889&-0.323587431 \\ 4.323587431 & -0.329608244& -5.231722584& 0.063001858\\ 4.260585573 &-0.014848109& -4.764684139& 0.003116284\\ 4.257469289 &-3.49722\times 10^{-05} & -4.742249655& 7.3746\times 10^{-06}\\ 4.257461914 &-1.95488\times 10^{-10}& -4.742196638&4.12231\times 10^{-11} \end {matrix}$

Therefore, $y \approx 4.257461914 \quad \Rightarrow x = y+2 \approx \boxed{6.26}$ .

Just graph it as 2 functions.