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

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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} $

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