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

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

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

We get rid of parentheses

3/5p-1p+40=0

We multiply all the terms by the denominator


-1p*5p+40*5p+3=0

Wy multiply elements

-5p^2+200p+3=0

a = -5; b = 200; c = +3;
Δ = b2-4ac
Δ = 2002-4·(-5)·3
Δ = 40060
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:

$\sqrt{\Delta}=\sqrt{40060}=\sqrt{4*10015}=\sqrt{4}*\sqrt{10015}=2\sqrt{10015}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-2\sqrt{10015}}{2*-5}=\frac{-200-2\sqrt{10015}}{-10}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+2\sqrt{10015}}{2*-5}=\frac{-200+2\sqrt{10015}}{-10}
$