M^3+m^2-42m=0

Solution for M^3+m^2-42m=0 equation:

^3+M^2-42M=0

We add all the numbers together, and all the variables

M^2-42M=0

a = 1; b = -42; c = 0;
Δ = b2-4ac
Δ = -422-4·1·0
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-42}{2*1}=\frac{0}{2}
=0
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+42}{2*1}=\frac{84}{2}
=42
$