1/5^2x-4=125

Solution for 1/5^2x-4=125 equation:

1/5^2x-4=125

We move all terms to the left:

1/5^2x-4-(125)=0


Domain of the equation: 5^2x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

1/5^2x-129=0

We multiply all the terms by the denominator


-129*5^2x+1=0

Wy multiply elements

-645x^2+1=0

a = -645; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-645)·1
Δ = 2580
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:

$\sqrt{\Delta}=\sqrt{2580}=\sqrt{4*645}=\sqrt{4}*\sqrt{645}=2\sqrt{645}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{645}}{2*-645}=\frac{0-2\sqrt{645}}{-1290}
=-\frac{2\sqrt{645}}{-1290}
=-\frac{\sqrt{645}}{-645}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{645}}{2*-645}=\frac{0+2\sqrt{645}}{-1290}
=\frac{2\sqrt{645}}{-1290}
=\frac{\sqrt{645}}{-645}
$