(1/4)=16^2x-5

Solution for (1/4)=16^2x-5 equation:

(1/4)=16^2x-5

We move all terms to the left:

(1/4)-(16^2x-5)=0

We add all the numbers together, and all the variables

-(16^2x-5)+(+1/4)=0

We get rid of parentheses

-16^2x+5+1/4=0

We multiply all the terms by the denominator


-16^2x*4+1+5*4=0

We add all the numbers together, and all the variables

-16^2x*4+21=0

Wy multiply elements

-64x^2+21=0

a = -64; b = 0; c = +21;
Δ = b2-4ac
Δ = 02-4·(-64)·21
Δ = 5376
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:

$\sqrt{\Delta}=\sqrt{5376}=\sqrt{256*21}=\sqrt{256}*\sqrt{21}=16\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{21}}{2*-64}=\frac{0-16\sqrt{21}}{-128}
=-\frac{16\sqrt{21}}{-128}
=-\frac{\sqrt{21}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{21}}{2*-64}=\frac{0+16\sqrt{21}}{-128}
=\frac{16\sqrt{21}}{-128}
=\frac{\sqrt{21}}{-8}
$