Решить уравнение (x^2+4x)^2+(x+2)^2 = 4

х^4+8х^3+16х^2+х^2+4х+4 = 4х^4+8х^3+17х^2+4х = 0х( х^3+8х^2+17х+4 ) = 0х( х^3+4х^2+4х^2+16х+х+4 ) = 0х( х^2(х+4)+4х(х+4)+1(х+4) ) = 0х(х+4)(х^2+4х+1) = 0х1 = 0х2 = - 4х^2+4х+1 = 0D = 16 - 4 = 12x3 = (-4+ 2 sqrt(3))/2 = - 2 + sqrt(3)x4 = (-4- 2 sqrt(3))/2 = - 2 - sqrt(3)