(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