Let \(u=2^x>0\). Then \(2^{x+2}=4u\) and \(2^{2-x}=4/u\), so

\[16u-\frac{20}{u}=3\quad\Longrightarrow\quad16u^2-3u-20=0.\]

\[u=\frac{3\pm\sqrt{1289}}{32}.\]

The minus sign gives a negative value, so reject it. Taking logarithms of the positive value,

\[x=\frac{\ln\left((3+\sqrt{1289})/32\right)}{\ln2}=0.281796\ldots=\boxed{0.282}\quad\text{(3 s.f.)}.\]