16-3p=(2/3p)+5

Solution for 16-3p=(2/3p)+5 equation:

16-3p=(2/3p)+5

We move all terms to the left:

16-3p-((2/3p)+5)=0


Domain of the equation: 3p)+5)!=0

p!=0/1

p!=0

p∈R


We add all the numbers together, and all the variables

-3p-((+2/3p)+5)+16=0

We multiply all the terms by the denominator


-3p*3p)+5)-((+16*3p)+5)+2=0

Wy multiply elements

-9p^2+48p=0

a = -9; b = 48; c = 0;
Δ = b2-4ac
Δ = 482-4·(-9)·0
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48}{2*-9}=\frac{-96}{-18}
=5+1/3
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48}{2*-9}=\frac{0}{-18}
=0
$