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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3n^2-10n-8=0

a = 3; b = -10; c = -8;
Δ = b2-4ac
Δ = -102-4·3·(-8)
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{-(-10)-14}{2*3}=\frac{-4}{6}
=-2/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+14}{2*3}=\frac{24}{6}
=4
$