(p-12)(p-6)=0

Solution for (p-12)(p-6)=0 equation:

(p-12)(p-6)=0

We multiply parentheses ..

(+p^2-6p-12p+72)=0

We get rid of parentheses

p^2-6p-12p+72=0

We add all the numbers together, and all the variables

p^2-18p+72=0

a = 1; b = -18; c = +72;
Δ = b2-4ac
Δ = -182-4·1·72
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6}{2*1}=\frac{12}{2}
=6
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6}{2*1}=\frac{24}{2}
=12
$