(r+5)(r+1)=0

Solution for (r+5)(r+1)=0 equation:

(r+5)(r+1)=0

We multiply parentheses ..

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

We get rid of parentheses

r^2+r+5r+5=0

We add all the numbers together, and all the variables

r^2+6r+5=0

a = 1; b = 6; c = +5;
Δ = b2-4ac
Δ = 62-4·1·5
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4}{2*1}=\frac{-10}{2}
=-5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4}{2*1}=\frac{-2}{2}
=-1
$