(4^x)+(2^x+3) = 20 С РЕШЕНИЕМ ПОМОГИТЕ, ПОЖАЛУЙСТА!!!!!!!!!!!!!!
0\\\\m^{2}+8m-20=0\\\\m_{1}=2\\\\m_{2}=-10" alt="4^{x}+2^{x+3}=20\\\\(2^{x})^{2}+2^{x}*2^{3} -20=0\\\\(2^{x})^{2}+8*2^{x}-20=0\\\\2^{x}=m,m>0\\\\m^{2}+8m-20=0\\\\m_{1}=2\\\\m_{2}=-10" align="absmiddle" class="latex-formula">