2(3^2x-1)=162

Solution for 2(3^2x-1)=162 equation:

2(3^2x-1)=162

We move all terms to the left:

2(3^2x-1)-(162)=0

We multiply parentheses

6x^2-2-162=0

We add all the numbers together, and all the variables

6x^2-164=0

a = 6; b = 0; c = -164;
Δ = b2-4ac
Δ = 02-4·6·(-164)
Δ = 3936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3936}=\sqrt{16*246}=\sqrt{16}*\sqrt{246}=4\sqrt{246}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{246}}{2*6}=\frac{0-4\sqrt{246}}{12}
=-\frac{4\sqrt{246}}{12}
=-\frac{\sqrt{246}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{246}}{2*6}=\frac{0+4\sqrt{246}}{12}
=\frac{4\sqrt{246}}{12}
=\frac{\sqrt{246}}{3}
$