(n+3)(2n+5)=0

Solution for (n+3)(2n+5)=0 equation:

(n+3)(2n+5)=0

We multiply parentheses ..

(+2n^2+5n+6n+15)=0

We get rid of parentheses

2n^2+5n+6n+15=0

We add all the numbers together, and all the variables

2n^2+11n+15=0

a = 2; b = 11; c = +15;
Δ = b2-4ac
Δ = 112-4·2·15
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{-(11)-1}{2*2}=\frac{-12}{4}
=-3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+1}{2*2}=\frac{-10}{4}
=-2+1/2
$