2/3p+1=-p-5

Solution for 2/3p+1=-p-5 equation:

2/3p+1=-p-5

We move all terms to the left:

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


Domain of the equation: 3p!=0

p!=0/3

p!=0

p∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

2/3p+1p+5+1=0

We multiply all the terms by the denominator


1p*3p+5*3p+1*3p+2=0

Wy multiply elements

3p^2+15p+3p+2=0

We add all the numbers together, and all the variables

3p^2+18p+2=0

a = 3; b = 18; c = +2;
Δ = b2-4ac
Δ = 182-4·3·2
Δ = 300
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:

$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-10\sqrt{3}}{2*3}=\frac{-18-10\sqrt{3}}{6}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+10\sqrt{3}}{2*3}=\frac{-18+10\sqrt{3}}{6}
$