(p-6)(p+3)=0

Solution for (p-6)(p+3)=0 equation:

(p-6)(p+3)=0

We multiply parentheses ..

(+p^2+3p-6p-18)=0

We get rid of parentheses

p^2+3p-6p-18=0

We add all the numbers together, and all the variables

p^2-3p-18=0

a = 1; b = -3; c = -18;
Δ = b2-4ac
Δ = -32-4·1·(-18)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-9}{2*1}=\frac{-6}{2}
=-3
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9}{2*1}=\frac{12}{2}
=6
$