(3n-8)(2n+4)=0

Solution for (3n-8)(2n+4)=0 equation:

(3n-8)(2n+4)=0

We multiply parentheses ..

(+6n^2+12n-16n-32)=0

We get rid of parentheses

6n^2+12n-16n-32=0

We add all the numbers together, and all the variables

6n^2-4n-32=0

a = 6; b = -4; c = -32;
Δ = b2-4ac
Δ = -42-4·6·(-32)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-28}{2*6}=\frac{-24}{12}
=-2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+28}{2*6}=\frac{32}{12}
=2+2/3
$