(х^2+2)^2-5(х^2+2)-6=0

Пусть (х²+2)=а, а∈R

a²-5a-6=0

D=(-(-5))²-4×1×(-6)=25+24=49

a1=(-(-5)-√49)/2×1=(5-7)/2=-2/2=-1

a2=(-(-5)+√49)/2×1=(5+7)/2=12/2=6

x²+2=a1

x²+2=-1

x²=-1-2

x²=-3

x1=-√3i=-1,73i

x2=√3i=1,73i

x²+2=a2

x²+2=6

x²=6-2

x²=4

x1=-√4

x1=-2

x2=√4

x2=2