4х^4-8х^3+3х^2-8х+4=0 С подробным объяснением

Решите задачу:

$4x^4-8x^3+3x^2-8x+4=0\\4(x^4+1)-8x(x^2+1)+3x^2=0\\4((x^2+1)^2-2x^2)-8x(x^2+1)+3x^2=0\\4(x^2+1)^2-8x^2-8x(x^2+1)+3x^2=0\\4(x^2+1)^2-8x(x^2+1)-5x^2=0\\ \frac{4(x^2+1)^2}{x^2} -\frac{8(x^2+1)}{x} -5=0\\\frac{x^2+1}{x} =t\\4t^2-8t-5=0\\D_1=16+20=36\\t_1=\frac{4+6}{4} =2,5\\t_2=-0,5\\\frac{x^2+1}{x} =-0,5\\2x^2+x+2=0\\D=-15\ \textless \ 0\\\frac{x^2+1}{x} =2,5\\2x^2-5x+2=0\\D=9\\x_1=2\\x_2=0,5$