Помогите решить уравнение:

$\displaystyle (8x^2 - 3x+1)^2 = 32x^2 - 12x+1\\ \\ (8x^2 - 3x+1)^2 -32x^2 +12x - 1 = 0\\ \\ \ z =8x^2 -3x \\ \\ -(4z+1) +(z^2 +2z+1)=0\\ \\ -4z-1+z^2 +2z+1=0\\ \\ -2z+z^2 =0\\ \\ -z *(2-z)=0\\ \\ -z = 0\\ \\ 2-z=0\\ \\ z = 0 \\ z=2\\$

$\displaystyle 8x -3 =0\\ \\ 8x^2-3x -2 =0 \\ \\ 1) x= 3/8\\ \\\\ 2)\ 8x^2-3x -2 \\ D= (-3)^2 - 4 *8*(-2) =73 \\ \\ x _1_,_2 =\frac{3б\sqrt{73}}{16} \\ \\$

ответ: x 1 = ( 3 - sqrt(73)) / 16 , x2 =0 , x3=3/8 , x4= (3+sqrt(73))/16