9^2+1=81^x-2/3x

Solution for 9^2+1=81^x-2/3x equation:

9^2+1=81^x-2/3x

We move all terms to the left:

9^2+1-(81^x-2/3x)=0


Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

-(81^x-2/3x)+82=0

We get rid of parentheses

-81^x+2/3x+82=0

We multiply all the terms by the denominator


-81^x*3x+82*3x+2=0

Wy multiply elements

-243x^2+246x+2=0

a = -243; b = 246; c = +2;
Δ = b2-4ac
Δ = 2462-4·(-243)·2
Δ = 62460
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:

$\sqrt{\Delta}=\sqrt{62460}=\sqrt{36*1735}=\sqrt{36}*\sqrt{1735}=6\sqrt{1735}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(246)-6\sqrt{1735}}{2*-243}=\frac{-246-6\sqrt{1735}}{-486}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(246)+6\sqrt{1735}}{2*-243}=\frac{-246+6\sqrt{1735}}{-486}
$