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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

12n^2-8n=0

a = 12; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·12·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*12}=\frac{0}{24}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*12}=\frac{16}{24}
=2/3
$