M3-3m2+4m-2=0

Solution for M3-3m2+4m-2=0 equation:

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

We add all the numbers together, and all the variables

-3M^2+4M+1=0

a = -3; b = 4; c = +1;
Δ = b2-4ac
Δ = 42-4·(-3)·1
Δ = 28
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{7}}{2*-3}=\frac{-4-2\sqrt{7}}{-6}
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{7}}{2*-3}=\frac{-4+2\sqrt{7}}{-6}
$