Question

x^3+4x^2- 24=0
Помогите)

Answer 1

$x^{3} +4 x^{2} -24=( x^{2} +6x+12)(x-2)=0$
1)
$x_{1}-2=0$
$x_{1} = 2$
2)
$x^{2} +6x+12=0$
$x_{2.3} = frac{-b+- sqrt{b^{2}-4ac} }{2a}= frac{-6+- sqrt{36-48} }{2} = frac{-6+-2 sqrt{-3} }{2} =-3+- isqrt{3}$
$x_{2} =-3+i sqrt{3}$
$x_{3} =-3-i sqrt{3}$