Answer:
[tex] {2}^{2x + 2} + 8 = 33( {2}^{x} ) \\ = > ( {2}^{2x} \times {2}^{2} ) + 8 = 33( {2}^{x} ) \\ let \: {2}^{x} be \: m \\ = > 4 {m}^{2} + 8 = 33m \\ = > 4 {m}^{2} - 33m + 8 = 0 \\ use \: quadratic \: equation \\ m = \frac{- b \frac{ + }{} \sqrt{( {b}^{2} - 4ac)} }{2a} [/tex]