(r+5)(r+13)=0

Solution for (r+5)(r+13)=0 equation:

(r+5)(r+13)=0

We multiply parentheses ..

(+r^2+13r+5r+65)=0

We get rid of parentheses

r^2+13r+5r+65=0

We add all the numbers together, and all the variables

r^2+18r+65=0

a = 1; b = 18; c = +65;
Δ = b2-4ac
Δ = 182-4·1·65
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-8}{2*1}=\frac{-26}{2}
=-13
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+8}{2*1}=\frac{-10}{2}
=-5
$