f(X)=2/4^

Solution for f(X)=2/4^ equation:

f(f)=2/4^

We move all terms to the left:

f(f)-(2/4^)=0

We get rid of parentheses

ff-2/4^=0

We multiply all the terms by the denominator


ff*4^-2=0

Wy multiply elements

4f^2-2=0

a = 4; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·4·(-2)
Δ = 32
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*4}=\frac{0-4\sqrt{2}}{8}
=-\frac{4\sqrt{2}}{8}
=-\frac{\sqrt{2}}{2}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*4}=\frac{0+4\sqrt{2}}{8}
=\frac{4\sqrt{2}}{8}
=\frac{\sqrt{2}}{2}
$