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∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

-9p^2-15p+48p-2=0

We add all the numbers together, and all the variables

-9p^2+33p-2=0

a = -9; b = 33; c = -2;
Δ = b2-4ac
Δ = 332-4·(-9)·(-2)
Δ = 1017
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:

$\sqrt{\Delta}=\sqrt{1017}=\sqrt{9*113}=\sqrt{9}*\sqrt{113}=3\sqrt{113}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-3\sqrt{113}}{2*-9}=\frac{-33-3\sqrt{113}}{-18}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+3\sqrt{113}}{2*-9}=\frac{-33+3\sqrt{113}}{-18}
$