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

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

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

We multiply parentheses ..

(+9n^2-15n+3n-5)=0

We get rid of parentheses

9n^2-15n+3n-5=0

We add all the numbers together, and all the variables

9n^2-12n-5=0

a = 9; b = -12; c = -5;
Δ = b2-4ac
Δ = -122-4·9·(-5)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-18}{2*9}=\frac{-6}{18}
=-1/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+18}{2*9}=\frac{30}{18}
=1+2/3
$