Z^3+3z^2+4z+12=0

Solution for Z^3+3z^2+4z+12=0 equation:

^3+3Z^2+4Z+12=0

We add all the numbers together, and all the variables

3Z^2+4Z=0

a = 3; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·3·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*3}=\frac{-8}{6}
=-1+1/3
$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*3}=\frac{0}{6}
=0
$