3/5p+.2(40-p)=0

Solution for 3/5p+.2(40-p)=0 equation:

3/5p+.2(40-p)=0


Domain of the equation: 5p!=0

p!=0/5

p!=0

p∈R


We add all the numbers together, and all the variables

3/5p+.2(-1p+40)=0

We multiply parentheses

3/5p-0.2p+8=0

We multiply all the terms by the denominator


-(0.2p)*5p+8*5p+3=0

We add all the numbers together, and all the variables

-(+0.2p)*5p+8*5p+3=0

We multiply parentheses

-0p^2+8*5p+3=0

Wy multiply elements

-0p^2+40p+3=0

We add all the numbers together, and all the variables

-1p^2+40p+3=0

a = -1; b = 40; c = +3;
Δ = b2-4ac
Δ = 402-4·(-1)·3
Δ = 1612
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1612}=\sqrt{4*403}=\sqrt{4}*\sqrt{403}=2\sqrt{403}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{403}}{2*-1}=\frac{-40-2\sqrt{403}}{-2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{403}}{2*-1}=\frac{-40+2\sqrt{403}}{-2}
$