Let

\[ u=2^x. \]

Then

\[ 2^{x+2}=4u \]

and

\[ 2^{2-x}=\frac{4}{u}. \]

So the equation becomes

\[ 4(4u)-5\left(\frac{4}{u}\right)=3. \]

Hence

\[ 16u-\frac{20}{u}=3. \]

Multiply by \(u\):

\[ 16u^2-20=3u. \]

So

\[ 16u^2-3u-20=0. \]

Using the quadratic formula,

\[ u=\frac{3\pm\sqrt{(-3)^2-4(16)(-20)}}{2(16)} =\frac{3\pm\sqrt{1289}}{32}. \]

Since \(u=2^x>0\),

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

Therefore

\[ 2^x=\frac{3+\sqrt{1289}}{32}. \]

Taking logarithms,

\[ x=\frac{\ln\left(\frac{3+\sqrt{1289}}{32}\right)}{\ln2}. \]

\[ x=0.281796\ldots \]

Answer: \(x=0.282\).