Z^3-5z^2-z+26=0

Solution for Z^3-5z^2-z+26=0 equation:

^3-5Z^2-Z+26=0

We add all the numbers together, and all the variables

-5Z^2-1Z=0

a = -5; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-5)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-5}=\frac{0}{-10}
=0
$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-5}=\frac{2}{-10}
=-1/5
$