Решите уравнение 4x^4+12x^3-47x^2+12x+4=0

$4x^4+12x^3-47x^2+12x+4=0 \\\ 4x^4-8x^3+20x^3-40x^2-7x^2+14x-2x+4=0 \\\ 4x^3(x-2)+20x^2(x-2)-7x(x-2)-2(x-2)=0 \\\ (x-2)(4x^3+20x^2-7x-2)=0 \\\ \boxed{x_1=2} \\\ 4x^3+20x^2-7x-2=0 \\\ 4x^3-2x^2+22x^2-11x+4x-2=0 \\\ 4x^2\left(x-\dfrac{1}{2}\right)+22x\left(x-\dfrac{1}{2}\right)+4\left(x-\dfrac{1}{2}\right)=0 \\\ \left(x-\dfrac{1}{2}\right)\left(4x^2+22x+4\right)=0 \\\ \boxed{x_2=\dfrac{1}{2}} \\\ 4x^2+22x+4=0 \\\ 2x^2+11x+2=0 \\\ D=11^2-4\cdot2\cdot2=121-16=105 \\\ \boxed {x_{34}=\dfrac{-11\pm\sqrt{105}}{4}}$
Ответ: 2; 1/2; $\dfrac{-11\pm\sqrt{105}}{4}$