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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

r^2-3r-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:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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