0.6p+0.2p(40-p)=0

Solution for 0.6p+0.2p(40-p)=0 equation:

0.6p+0.2p(40-p)=0

We add all the numbers together, and all the variables

0.6p+0.2p(-1p+40)=0

We multiply parentheses

0p^2+0.6p+0p=0

We add all the numbers together, and all the variables

p^2+1.6p=0

a = 1; b = 1.6; c = 0;
Δ = b2-4ac
Δ = 1.62-4·1·0
Δ = 2.56
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1.6)-\sqrt{2.56}}{2*1}=\frac{-1.6-\sqrt{2.56}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.6)+\sqrt{2.56}}{2*1}=\frac{-1.6+\sqrt{2.56}}{2}
$