M^3+3m^2-4m+12=0

Solution for M^3+3m^2-4m+12=0 equation:

^3+3M^2-4M+12=0

We add all the numbers together, and all the variables

3M^2-4M=0

a = 3; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·3·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*3}=\frac{0}{6}
=0
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*3}=\frac{8}{6}
=1+1/3
$