7^2^x^2^-x=1/7

Solution for 7^2^x^2^-x=1/7 equation:

7^2^x^2^-x=1/7

We move all terms to the left:

7^2^x^2^-x-(1/7)=0

We add all the numbers together, and all the variables

7^2^x^2^-x-(+1/7)=0

We add all the numbers together, and all the variables

-1x+7^2^x^2^-(+1/7)=0

We get rid of parentheses

-1x+7^2^x^2^-1/7=0

We multiply all the terms by the denominator


-1x*7+7^2^x^2^*7-1=0

Wy multiply elements

49x^2-7x-1=0

a = 49; b = -7; c = -1;
Δ = b2-4ac
Δ = -72-4·49·(-1)
Δ = 245
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:

$\sqrt{\Delta}=\sqrt{245}=\sqrt{49*5}=\sqrt{49}*\sqrt{5}=7\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7\sqrt{5}}{2*49}=\frac{7-7\sqrt{5}}{98}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7\sqrt{5}}{2*49}=\frac{7+7\sqrt{5}}{98}
$