2^2x=1/16

Solution for 2^2x=1/16 equation:

2^2x=1/16

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

2^2x-1/16=0

We multiply all the terms by the denominator


2^2x*16-1=0

Wy multiply elements

32x^2-1=0

a = 32; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·32·(-1)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{2}}{2*32}=\frac{0-8\sqrt{2}}{64}
=-\frac{8\sqrt{2}}{64}
=-\frac{\sqrt{2}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*32}=\frac{0+8\sqrt{2}}{64}
=\frac{8\sqrt{2}}{64}
=\frac{\sqrt{2}}{8}
$