(-m-3)(m+12)=0

Solution for (-m-3)(m+12)=0 equation:

(-m-3)(m+12)=0

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-1m^2-12m-3m-36)=0

We get rid of parentheses

-1m^2-12m-3m-36=0

We add all the numbers together, and all the variables

-1m^2-15m-36=0

a = -1; b = -15; c = -36;
Δ = b2-4ac
Δ = -152-4·(-1)·(-36)
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*-1}=\frac{6}{-2}
=-3
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+9}{2*-1}=\frac{24}{-2}
=-12
$