6(3p^2+p)=0

Solution for 6(3p^2+p)=0 equation:

6(3p^2+p)=0

We multiply parentheses

18p^2+6p=0

a = 18; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·18·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*18}=\frac{-12}{36}
=-1/3
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*18}=\frac{0}{36}
=0
$