A^3+a^2-36a=36

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

^3+A^2-36A=36

We move all terms to the left:

^3+A^2-36A-(36)=0

determiningTheFunctionDomain
A^2-36A-36+^3=0

We add all the numbers together, and all the variables

A^2-36A=0

a = 1; b = -36; c = 0;
Δ = b2-4ac
Δ = -362-4·1·0
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-36}{2*1}=\frac{0}{2}
=0
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+36}{2*1}=\frac{72}{2}
=36
$