3p^2-39p+120=0

Solution for 3p^2-39p+120=0 equation:

3p^2-39p+120=0

a = 3; b = -39; c = +120;
Δ = b2-4ac
Δ = -392-4·3·120
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-9}{2*3}=\frac{30}{6}
=5
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+9}{2*3}=\frac{48}{6}
=8
$