14-6p(3p+1)=(4+p)

Solution for 14-6p(3p+1)=(4+p) equation:

14-6p(3p+1)=(4+p)

We move all terms to the left:

14-6p(3p+1)-((4+p))=0

We add all the numbers together, and all the variables

-6p(3p+1)-((p+4))+14=0

We multiply parentheses

-18p^2-6p-((p+4))+14=0


We calculate terms in parentheses: -((p+4)), so:

(p+4)

We get rid of parentheses

p+4

Back to the equation:

-(p+4)


We get rid of parentheses

-18p^2-6p-p-4+14=0

We add all the numbers together, and all the variables

-18p^2-7p+10=0

a = -18; b = -7; c = +10;
Δ = b2-4ac
Δ = -72-4·(-18)·10
Δ = 769
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{-(-7)-\sqrt{769}}{2*-18}=\frac{7-\sqrt{769}}{-36}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{769}}{2*-18}=\frac{7+\sqrt{769}}{-36}
$