[tex]\displaystyle\\4^x-6\cdot2^x+8=0\\\\\Big(2^2\Big)^x-6\cdot2^x+8=0\\\\2^{2x}-6\cdot2^x+8=0\\\\\Big(2^x\Big)^2-6\cdot2^x+8=0\\\\\text{Facem substitutia: }\boxed{2^x=y}\\\\y^2-6y+8=0\\\\y_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{6\pm\sqrt{(-6)^2-4\cdot1\cdot8}}{2\cdot1}=\\\\=\frac{6\pm\sqrt{36-32}}{2}=\frac{6\pm\sqrt{4}}{2}=\frac{6\pm2}{2}=3\pm1\\\\y_1=3-1=\boxed{2}\\y_2=3+1=\boxed{4}\\\text{Ne intoarcem la substitutie.}\\2^x=y_1\\2^x=2=2^1\\\boxed{x_1=1}\\\\2^x=y_2\\2^x=4=2^2\\\boxed{x_2=2}[/tex]
