Question

x^4+12x+3=0. Найдите x

Answer 1

$x^4+12x+3=0$

$x^{2}=t$

$t^{2}+12t+3=0$

$D=b^{2}-4ac=12^{2}-4*1*3=132$

$t=\frac{-b-\sqrt{D} }{2a}=\frac{-12-\sqrt{132} }{2}=-11.745$

$t=\frac{-b+\sqrt{D} }{2a}=\frac{-12+\sqrt{132} }{2}=-0.255$

Корней нет!