(M-3)(m+4)(m+1)=0

Solution for (M-3)(m+4)(m+1)=0 equation:

(-3)(M+4)(M+1)=0

We multiply parentheses ..

(-3M-12)(M+1)=0

We multiply parentheses ..

(-3M^2-3M-12M-12)=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3M^2-15M-12=0

a = -3; b = -15; c = -12;
Δ = b2-4ac
Δ = -152-4·(-3)·(-12)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-9}{2*-3}=\frac{6}{-6}
=-1
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+9}{2*-3}=\frac{24}{-6}
=-4
$