(9x+3-6^x)(3x)=1

Solution for (9x+3-6^x)(3x)=1 equation:

(9x+3-6^x)(3x)=1

We move all terms to the left:

(9x+3-6^x)(3x)-(1)=0

We multiply parentheses

27x^2-18x^2+9x-1=0

We add all the numbers together, and all the variables

9x^2+9x-1=0

a = 9; b = 9; c = -1;
Δ = b2-4ac
Δ = 92-4·9·(-1)
Δ = 117
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:

$\sqrt{\Delta}=\sqrt{117}=\sqrt{9*13}=\sqrt{9}*\sqrt{13}=3\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{13}}{2*9}=\frac{-9-3\sqrt{13}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{13}}{2*9}=\frac{-9+3\sqrt{13}}{18}
$