16-3p=2/3p+

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

16-3p=2/3p+

We move all terms to the left:

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


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

p∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

-3p-2/3p+16=0

We multiply all the terms by the denominator


-3p*3p+16*3p-2=0

Wy multiply elements

-9p^2+48p-2=0

a = -9; b = 48; c = -2;
Δ = b2-4ac
Δ = 482-4·(-9)·(-2)
Δ = 2232
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:

$\sqrt{\Delta}=\sqrt{2232}=\sqrt{36*62}=\sqrt{36}*\sqrt{62}=6\sqrt{62}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-6\sqrt{62}}{2*-9}=\frac{-48-6\sqrt{62}}{-18}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+6\sqrt{62}}{2*-9}=\frac{-48+6\sqrt{62}}{-18}
$