(n+4)(n+11)=0

Solution for (n+4)(n+11)=0 equation:

(n+4)(n+11)=0

We multiply parentheses ..

(+n^2+11n+4n+44)=0

We get rid of parentheses

n^2+11n+4n+44=0

We add all the numbers together, and all the variables

n^2+15n+44=0

a = 1; b = 15; c = +44;
Δ = b2-4ac
Δ = 152-4·1·44
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-7}{2*1}=\frac{-22}{2}
=-11
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+7}{2*1}=\frac{-8}{2}
=-4
$