3n^2+20n+32=0

Solution for 3n^2+20n+32=0 equation:

3n^2+20n+32=0

a = 3; b = 20; c = +32;
Δ = b2-4ac
Δ = 202-4·3·32
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{-(20)-4}{2*3}=\frac{-24}{6}
=-4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4}{2*3}=\frac{-16}{6}
=-2+2/3
$