Răspuns:
Explicație pas cu pas:
[tex]4^x-2^x=12\\ \\ (2^2)^x-2^x=12 \\ \\ 2^{2x}-2^{x}=12\\ \\ (2^{x})^2-2^x=12\\ \\ Notez~2^x=t~sau~orice~alta~litera\\ \\ t^2-t-12=0 \\ \\ \Delta=(-1)^2-4\cdot1\cdot(-12)=1+48=49 \Rightarrow \sqrt{\Delta}=7\\ \\ t_{1}=\frac{1+7}{2}=4\\ \\ t_2=\frac{1-7}{2}=-3,~imposibil~ca~2^x=-3,~deoarece~2>0\\ \\Deci~2^x=4 \Rightarrow 2^x=2^2 \\ \Rightarrow \boxed{x=2}[/tex]
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[tex]\boxed{Ecuatia:~ax^2+bx+c=0 \\ \\ Are~solutiile~x_{1,2}=\frac{-b+-\sqrt{\Delta}}{2a},~unde~\Delta=b^2-4ac}[/tex]
