(n+3)(n+4)=0

Solution for (n+3)(n+4)=0 equation:

(n+3)(n+4)=0

We multiply parentheses ..

(+n^2+4n+3n+12)=0

We get rid of parentheses

n^2+4n+3n+12=0

We add all the numbers together, and all the variables

n^2+7n+12=0

a = 1; b = 7; c = +12;
Δ = b2-4ac
Δ = 72-4·1·12
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*1}=\frac{-8}{2}
=-4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*1}=\frac{-6}{2}
=-3
$