Z^3+-9z^2+14z+24=0

Solution for Z^3+-9z^2+14z+24=0 equation:

^3+-9Z^2+14Z+24=0

We add all the numbers together, and all the variables

-9Z^2+14Z=0

a = -9; b = 14; c = 0;
Δ = b2-4ac
Δ = 142-4·(-9)·0
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14}{2*-9}=\frac{-28}{-18}
=1+5/9
$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14}{2*-9}=\frac{0}{-18}
=0
$