[tex]\displaystyle Notam~3^x=t\ \textgreater \ 0. \\ \\ Ecuatia~se~rescrie~sub~forma~echivalenta: \boxed{ t^2+t-12=0}~. \\ \\ \Delta=1-4 \cdot 1 \cdot (-12)=49. \\ \\ t_1= \frac{-1+ \sqrt{49}}{2}= \frac{-1+7}{2}=3. \\ \\ t_2= \frac{-1- \sqrt{49}}{2}= \frac{-1-7}{2}=-4~(nu~convine,~caci~3^x=t\ \textgreater \ 0.) \\ \\ Deci~3^x=3 \Rightarrow \boxed{x=1}~. \\ \\ Solutie:~x=1.\\ \\ Observatie:~ecuatia~din~enunt~se~putea~rescrie~astfel: \\ \\ (3^x-3)(3^x+4)=0.[/tex]
