3^2x-3=27^x+1/3

Solution for 3^2x-3=27^x+1/3 equation:

3^2x-3=27^x+1/3

We move all terms to the left:

3^2x-3-(27^x+1/3)=0

We get rid of parentheses

3^2x-27^x-3-1/3=0

We multiply all the terms by the denominator


3^2x*3-27^x*3-1-3*3=0

We add all the numbers together, and all the variables

3^2x*3-27^x*3-10=0

Wy multiply elements

9x^2-81x-10=0

a = 9; b = -81; c = -10;
Δ = b2-4ac
Δ = -812-4·9·(-10)
Δ = 6921
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:

$\sqrt{\Delta}=\sqrt{6921}=\sqrt{9*769}=\sqrt{9}*\sqrt{769}=3\sqrt{769}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-81)-3\sqrt{769}}{2*9}=\frac{81-3\sqrt{769}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-81)+3\sqrt{769}}{2*9}=\frac{81+3\sqrt{769}}{18}
$