3(3^2p-6)=19

Solution for 3(3^2p-6)=19 equation:

3(3^2p-6)=19

We move all terms to the left:

3(3^2p-6)-(19)=0

We multiply parentheses

9p^2-18-19=0

We add all the numbers together, and all the variables

9p^2-37=0

a = 9; b = 0; c = -37;
Δ = b2-4ac
Δ = 02-4·9·(-37)
Δ = 1332
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:

$\sqrt{\Delta}=\sqrt{1332}=\sqrt{36*37}=\sqrt{36}*\sqrt{37}=6\sqrt{37}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{37}}{2*9}=\frac{0-6\sqrt{37}}{18}
=-\frac{6\sqrt{37}}{18}
=-\frac{\sqrt{37}}{3}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{37}}{2*9}=\frac{0+6\sqrt{37}}{18}
=\frac{6\sqrt{37}}{18}
=\frac{\sqrt{37}}{3}
$