7/2p+1-3/p+1=2/p+1

Solution for 7/2p+1-3/p+1=2/p+1 equation:

7/2p+1-3/p+1=2/p+1

We move all terms to the left:

7/2p+1-3/p+1-(2/p+1)=0


Domain of the equation: 2p!=0

p!=0/2

p!=0

p∈R



Domain of the equation: p!=0

p∈R



Domain of the equation: p+1)!=0

p∈R


We add all the numbers together, and all the variables

7/2p-3/p-(2/p+1)+2=0

We get rid of parentheses

7/2p-3/p-2/p-1+2=0

We calculate fractions


7p/2p^2+(-4p-3)/2p^2-1+2=0

We add all the numbers together, and all the variables

7p/2p^2+(-4p-3)/2p^2+1=0

We multiply all the terms by the denominator


7p+(-4p-3)+1*2p^2=0

Wy multiply elements

2p^2+7p+(-4p-3)=0

We get rid of parentheses

2p^2+7p-4p-3=0

We add all the numbers together, and all the variables

2p^2+3p-3=0

a = 2; b = 3; c = -3;
Δ = b2-4ac
Δ = 32-4·2·(-3)
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{33}}{2*2}=\frac{-3-\sqrt{33}}{4}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{33}}{2*2}=\frac{-3+\sqrt{33}}{4}
$