4-2f(f=1)

Solution for 4-2f(f=1) equation:

4-2f(f=1)

We move all terms to the left:

4-2f(f-(1))=0

We multiply parentheses

-2f^2+2f+4=0

a = -2; b = 2; c = +4;
Δ = b2-4ac
Δ = 22-4·(-2)·4
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*-2}=\frac{-8}{-4}
=+2
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*-2}=\frac{4}{-4}
=-1
$