[tex]\displaystyle\bf\\4^x+2^x-2=0\\\\\Big(2^2\Big)^x+2^x-2=0\\\\\Big(2^x\Big)^2+2^x-2=0\\\\\boxed{\bf2^x=y}~~~(Substitutie)\\\\y^2+y-2=0\\\\y_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1+8}}{2}=\frac{-1\pm\sqrt{9}}{2}=\frac{-1\pm3}{2}\\\\y_1=\frac{-1-3}{2}=\frac{-4}{2}=-2\\\\y_2=\frac{-1+3}{2}=\frac{2}{2}=1\\\\Ne~intoarcem~la~substitutie.\\\\2^x=-2~~~Nu~are~solutie~reala\\\\2^x=1\\\\\implies~~\boxed{\bf x=0}~~Solutie~unica[/tex]
