2(n+5)(n+1)=0

Solution for 2(n+5)(n+1)=0 equation:

2(n+5)(n+1)=0

We multiply parentheses ..

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

We multiply parentheses

2n^2+2n+10n+10=0

We add all the numbers together, and all the variables

2n^2+12n+10=0

a = 2; b = 12; c = +10;
Δ = b2-4ac
Δ = 122-4·2·10
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8}{2*2}=\frac{-20}{4}
=-5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8}{2*2}=\frac{-4}{4}
=-1
$