2/3p-1=1+p

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

2/3p-1=1+p

We move all terms to the left:

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

We get rid of parentheses

2/3p-p-1-1=0

We multiply all the terms by the denominator


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

Wy multiply elements

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

We add all the numbers together, and all the variables

-3p^2-6p+2=0

a = -3; b = -6; c = +2;
Δ = b2-4ac
Δ = -62-4·(-3)·2
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{15}}{2*-3}=\frac{6-2\sqrt{15}}{-6}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{15}}{2*-3}=\frac{6+2\sqrt{15}}{-6}
$