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 \neq -0,25; \\ 2^x = 1; 2^x = 2^0; x = 0 [/latex]