2(3p+2)-3(2p-3)p=2

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

2(3p+2)-3(2p-3)p=2

We move all terms to the left:

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

We multiply parentheses

-6p^2+6p+9p+4-2=0

We add all the numbers together, and all the variables

-6p^2+15p+2=0

a = -6; b = 15; c = +2;
Δ = b2-4ac
Δ = 152-4·(-6)·2
Δ = 273
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{273}}{2*-6}=\frac{-15-\sqrt{273}}{-12}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{273}}{2*-6}=\frac{-15+\sqrt{273}}{-12}
$