(p-6)(p-2)=0

Solution for (p-6)(p-2)=0 equation:

(p-6)(p-2)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

p^2-8p+12=0

a = 1; b = -8; c = +12;
Δ = b2-4ac
Δ = -82-4·1·12
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*1}=\frac{4}{2}
=2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*1}=\frac{12}{2}
=6
$