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

Solution for (2n+3)(n+4)=0 equation:

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

We multiply parentheses ..

(+2n^2+8n+3n+12)=0

We get rid of parentheses

2n^2+8n+3n+12=0

We add all the numbers together, and all the variables

2n^2+11n+12=0

a = 2; b = 11; c = +12;
Δ = b2-4ac
Δ = 112-4·2·12
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-5}{2*2}=\frac{-16}{4}
=-4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+5}{2*2}=\frac{-6}{4}
=-1+1/2
$