(r+2)(r+3)=0

Solution for (r+2)(r+3)=0 equation:

(r+2)(r+3)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

r^2+5r+6=0

a = 1; b = 5; c = +6;
Δ = b2-4ac
Δ = 52-4·1·6
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-1}{2*1}=\frac{-6}{2}
=-3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*1}=\frac{-4}{2}
=-2
$