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

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

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

We multiply parentheses ..

-(+r^2+3r-12r-36)=0

We get rid of parentheses

-r^2-3r+12r+36=0

We add all the numbers together, and all the variables

-1r^2+9r+36=0

a = -1; b = 9; c = +36;
Δ = b2-4ac
Δ = 92-4·(-1)·36
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-15}{2*-1}=\frac{-24}{-2}
=+12
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+15}{2*-1}=\frac{6}{-2}
=-3
$