(n+4)(n+7)=0

Solution for (n+4)(n+7)=0 equation:

(n+4)(n+7)=0

We multiply parentheses ..

(+n^2+7n+4n+28)=0

We get rid of parentheses

n^2+7n+4n+28=0

We add all the numbers together, and all the variables

n^2+11n+28=0

a = 1; b = 11; c = +28;
Δ = b2-4ac
Δ = 112-4·1·28
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-3}{2*1}=\frac{-14}{2}
=-7
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+3}{2*1}=\frac{-8}{2}
=-4
$