(r+3)(r+13)=0

Solution for (r+3)(r+13)=0 equation:

(r+3)(r+13)=0

We multiply parentheses ..

(+r^2+13r+3r+39)=0

We get rid of parentheses

r^2+13r+3r+39=0

We add all the numbers together, and all the variables

r^2+16r+39=0

a = 1; b = 16; c = +39;
Δ = b2-4ac
Δ = 162-4·1·39
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-10}{2*1}=\frac{-26}{2}
=-13
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+10}{2*1}=\frac{-6}{2}
=-3
$