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Question Number 157453 by MathSh last updated on 23/Oct/21

Answered by MJS_new last updated on 23/Oct/21

(x+2)(x^2 +1)(x^2 +x+6)(x^2 −x−1)=0  x=−2 ∨ x=((1±(√5))/2) ∨ x=±i ∨ x=−((1±(√(23))i)/2)

$$\left({x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{6}\right)\left({x}^{\mathrm{2}} −{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{2}\:\vee\:{x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\:\vee\:{x}=\pm\mathrm{i}\:\vee\:{x}=−\frac{\mathrm{1}\pm\sqrt{\mathrm{23}}\mathrm{i}}{\mathrm{2}} \\ $$

Commented by MathSh last updated on 23/Oct/21

Thank you so much dear Ser  Can you specify where it came from in  detail

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{dear}\:\boldsymbol{\mathrm{Ser}} \\ $$$$\mathrm{Can}\:\mathrm{you}\:\mathrm{specify}\:\mathrm{where}\:\mathrm{it}\:\mathrm{came}\:\mathrm{from}\:\mathrm{in} \\ $$$$\mathrm{detail} \\ $$

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