\(4(2^{x^2}) = 8^{x}  \equiv (2^{2})(2^{x^2}) = (2^{3})^{x}\)

\(\implies 2^{2 + x^{2}} = 2^{3x}\)

Comparing bases, we have

\(2 + x^{2} = 3x \implies x^{2} - 3x + 2 = 0\)

\(x^{2} - 2x - x + 2 = 0 \)

\(x(x - 2) - 1(x - 2) = 0\)

\((x - 1) = 0\) or \((x - 2) = 0\)

\(x = \text{1 or 2}\)