(2p+3)(3p+1)=5

Solution for (2p+3)(3p+1)=5 equation:

(2p+3)(3p+1)=5

We move all terms to the left:

(2p+3)(3p+1)-(5)=0

We multiply parentheses ..

(+6p^2+2p+9p+3)-5=0

We get rid of parentheses

6p^2+2p+9p+3-5=0

We add all the numbers together, and all the variables

6p^2+11p-2=0

a = 6; b = 11; c = -2;
Δ = b2-4ac
Δ = 112-4·6·(-2)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-13}{2*6}=\frac{-24}{12}
=-2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+13}{2*6}=\frac{2}{12}
=1/6
$