-3n(4n-5)=0

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

-3n(4n-5)=0

We multiply parentheses

-12n^2+15n=0

a = -12; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·(-12)·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*-12}=\frac{-30}{-24}
=1+1/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*-12}=\frac{0}{-24}
=0
$