3(3^2p-6)=27

Solution for 3(3^2p-6)=27 equation:

3(3^2p-6)=27

We move all terms to the left:

3(3^2p-6)-(27)=0

We multiply parentheses

9p^2-18-27=0

We add all the numbers together, and all the variables

9p^2-45=0

a = 9; b = 0; c = -45;
Δ = b2-4ac
Δ = 02-4·9·(-45)
Δ = 1620
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:

$\sqrt{\Delta}=\sqrt{1620}=\sqrt{324*5}=\sqrt{324}*\sqrt{5}=18\sqrt{5}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{5}}{2*9}=\frac{0-18\sqrt{5}}{18}
=-\frac{18\sqrt{5}}{18}
=-\sqrt{5}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{5}}{2*9}=\frac{0+18\sqrt{5}}{18}
=\frac{18\sqrt{5}}{18}
=\sqrt{5}
$