(2n-4)(3n+5)=0

Solution for (2n-4)(3n+5)=0 equation:

(2n-4)(3n+5)=0

We multiply parentheses ..

(+6n^2+10n-12n-20)=0

We get rid of parentheses

6n^2+10n-12n-20=0

We add all the numbers together, and all the variables

6n^2-2n-20=0

a = 6; b = -2; c = -20;
Δ = b2-4ac
Δ = -22-4·6·(-20)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-22}{2*6}=\frac{-20}{12}
=-1+2/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+22}{2*6}=\frac{24}{12}
=2
$