(c+9)(c-12)=0

Solution for (c+9)(c-12)=0 equation:

(c+9)(c-12)=0

We multiply parentheses ..

(+c^2-12c+9c-108)=0

We get rid of parentheses

c^2-12c+9c-108=0

We add all the numbers together, and all the variables

c^2-3c-108=0

a = 1; b = -3; c = -108;
Δ = b2-4ac
Δ = -32-4·1·(-108)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-21}{2*1}=\frac{-18}{2}
=-9
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+21}{2*1}=\frac{24}{2}
=12
$