Помогите решить уравнения:

1)
3x - 4 = 5x-1
-2x = 3
x = $\frac{-3}{2}$

2)
$log_{2}(x^2+4x + 4) = log_{2}16 \\ x^2+4x + 4 = 16 \\ x^2+4x -12 = 0 \\ D = 16 + 48 = 64\\ x_1 = \frac{-4+8}{2} = 2\\ x_2 = \frac{-4-8}{2} = -6\\$

3)
$lg^2x + lgx - 2 = 0\\ lgx = t\\ t^2 + t - 2 = 0\\D = 1 + 8 = 9\\x_1 = \frac{-1+3}{2} = 1\\ x_2 = \frac{-1-3}{2} = -2\\ (1) lgx = 1 \\ x = 10 \\ (2)lgx = -2\\x = 10^{-2}$