2/3p+6=p+12

Solution for 2/3p+6=p+12 equation:

2/3p+6=p+12

We move all terms to the left:

2/3p+6-(p+12)=0


Domain of the equation: 3p!=0

p!=0/3

p!=0

p∈R


We get rid of parentheses

2/3p-p-12+6=0

We multiply all the terms by the denominator


-p*3p-12*3p+6*3p+2=0

Wy multiply elements

-3p^2-36p+18p+2=0

We add all the numbers together, and all the variables

-3p^2-18p+2=0

a = -3; b = -18; c = +2;
Δ = b2-4ac
Δ = -182-4·(-3)·2
Δ = 348
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:

$\sqrt{\Delta}=\sqrt{348}=\sqrt{4*87}=\sqrt{4}*\sqrt{87}=2\sqrt{87}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{87}}{2*-3}=\frac{18-2\sqrt{87}}{-6}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{87}}{2*-3}=\frac{18+2\sqrt{87}}{-6}
$