Ребят помогите 4^[х+1]-3×2^[х]-1=0

[latex] 4^{x+1}-3*2^{x}-1=0 \2^{2(x+1}-3*2^{x}-1=0\2^{2x+2}-3*2^{x}-1=0\2^2*2^{2x}-3*2^{x}-1=0 \4*2^{2x}-3*2^{x}-1=0 \2^x = t \4t^2-3t-1 = 0 \D = (-3)^2+4*4*1=9+16=25 \t_{1,2} = frac{3б5}{8} \t_1 = 1; t_2 = -frac{2}{8} = -0,25 \2^x eq -0,25; \2^x = 1; 2^x = 2^0; x = 0 [/latex]