f^2-5=4f

Solution for f^2-5=4f equation:

f^2-5=4f

We move all terms to the left:

f^2-5-(4f)=0

a = 1; b = -4; c = -5;
Δ = b2-4ac
Δ = -42-4·1·(-5)
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{-(-4)-6}{2*1}=\frac{-2}{2}
=-1
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6}{2*1}=\frac{10}{2}
=5
$