3(n^2)-16n+20=0

Solution for 3(n^2)-16n+20=0 equation:

3(n^2)-16n+20=0

a = 3; b = -16; c = +20;
Δ = b2-4ac
Δ = -162-4·3·20
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4}{2*3}=\frac{12}{6}
=2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4}{2*3}=\frac{20}{6}
=3+1/3
$