18-3(2p+4)3p=3(2-3p)

Solution for 18-3(2p+4)3p=3(2-3p) equation:

18-3(2p+4)3p=3(2-3p)

We move all terms to the left:

18-3(2p+4)3p-(3(2-3p))=0

We add all the numbers together, and all the variables

-3(2p+4)3p-(3(-3p+2))+18=0

We multiply parentheses

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


We calculate terms in parentheses: -(3(-3p+2)), so:

3(-3p+2)

We multiply parentheses

-9p+6

Back to the equation:

-(-9p+6)


We get rid of parentheses

-18p^2-36p+9p-6+18=0

We add all the numbers together, and all the variables

-18p^2-27p+12=0

a = -18; b = -27; c = +12;
Δ = b2-4ac
Δ = -272-4·(-18)·12
Δ = 1593
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:

$\sqrt{\Delta}=\sqrt{1593}=\sqrt{9*177}=\sqrt{9}*\sqrt{177}=3\sqrt{177}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-3\sqrt{177}}{2*-18}=\frac{27-3\sqrt{177}}{-36}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+3\sqrt{177}}{2*-18}=\frac{27+3\sqrt{177}}{-36}
$