(3n+1)(n+5)=0

Solution for (3n+1)(n+5)=0 equation:

(3n+1)(n+5)=0

We multiply parentheses ..

(+3n^2+15n+n+5)=0

We get rid of parentheses

3n^2+15n+n+5=0

We add all the numbers together, and all the variables

3n^2+16n+5=0

a = 3; b = 16; c = +5;
Δ = b2-4ac
Δ = 162-4·3·5
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-14}{2*3}=\frac{-30}{6}
=-5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+14}{2*3}=\frac{-2}{6}
=-1/3
$