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Ответ

Ответ дан
viktorz2014
 1x^{3}+5 x^{2}+0x-5=0
Коэффициенты: a=5, b=0, c=-5

Q= frac{ a^{2}-3b }{9} = frac{5 ^{2}-3*0 }{9} =2.77778
R= frac{2a ^{3}-9ab+27c }{54} = frac{2*5 ^{3}-9*5*0+27*(-5) }{54} =2.12963
S=Q ^{3}-R ^{2}  =16.89815
Если S>0 вычисляем φ= frac{1}{3} arccos( frac{R}{ sqrt{Q ^{3} } } )
φ= frac{1}{3}arccos( frac{2.212963}{ sqrt{2.77778 ^{3} } } ) =0.36427
и имеем три действительных корня:

x₁=-2 sqrt{Q} cosφ - frac{a}{3} =-4.781
x₂=-2 sqrt{Q} cos(φ+ frac{2}{3}  pi )- frac{a}{3} =0.919
x₃=-2 sqrt{Q}cos (φ- frac{2}{3}  pi - frac{a}{3} =-1.138

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