2n(3n-5)=4

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

2n(3n-5)=4

We move all terms to the left:

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

We multiply parentheses

6n^2-10n-4=0

a = 6; b = -10; c = -4;
Δ = b2-4ac
Δ = -102-4·6·(-4)
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*6}=\frac{-4}{12}
=-1/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+14}{2*6}=\frac{24}{12}
=2
$