(3n-1)(n-4)=0

Solution for (3n-1)(n-4)=0 equation:

(3n-1)(n-4)=0

We multiply parentheses ..

(+3n^2-12n-1n+4)=0

We get rid of parentheses

3n^2-12n-1n+4=0

We add all the numbers together, and all the variables

3n^2-13n+4=0

a = 3; b = -13; c = +4;
Δ = b2-4ac
Δ = -132-4·3·4
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-11}{2*3}=\frac{2}{6}
=1/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+11}{2*3}=\frac{24}{6}
=4
$