A^3-5a^2-6a+36=0

Solution for A^3-5a^2-6a+36=0 equation:

^3-5A^2-6A+36=0

We add all the numbers together, and all the variables

-5A^2-6A=0

a = -5; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·(-5)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*-5}=\frac{0}{-10}
=0
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*-5}=\frac{12}{-10}
=-1+1/5
$