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

p+3/5p+40=0

We multiply all the terms by the denominator


p*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 = 39940
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:

$\sqrt{\Delta}=\sqrt{39940}=\sqrt{4*9985}=\sqrt{4}*\sqrt{9985}=2\sqrt{9985}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-2\sqrt{9985}}{2*5}=\frac{-200-2\sqrt{9985}}{10}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+2\sqrt{9985}}{2*5}=\frac{-200+2\sqrt{9985}}{10}
$