-4n(4n-3)=0

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

-4n(4n-3)=0

We multiply parentheses

-16n^2+12n=0

a = -16; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-16)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*-16}=\frac{-24}{-32}
=3/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-16}=\frac{0}{-32}
=0
$