2^x*3^x=216^2

Solution for 2^x*3^x=216^2 equation:

2^x*3^x=216^2

We move all terms to the left:

2^x*3^x-(216^2)=0

We add all the numbers together, and all the variables

2^x*3^x-46656=0

Wy multiply elements

6x^2-46656=0

a = 6; b = 0; c = -46656;
Δ = b2-4ac
Δ = 02-4·6·(-46656)
Δ = 1119744
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:

$\sqrt{\Delta}=\sqrt{1119744}=\sqrt{186624*6}=\sqrt{186624}*\sqrt{6}=432\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-432\sqrt{6}}{2*6}=\frac{0-432\sqrt{6}}{12}
=-\frac{432\sqrt{6}}{12}
=-36\sqrt{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+432\sqrt{6}}{2*6}=\frac{0+432\sqrt{6}}{12}
=\frac{432\sqrt{6}}{12}
=36\sqrt{6}
$