F(c)=c2-4c

Solution for F(c)=c2-4c equation:

(F)=F2-4F

We move all terms to the left:

(F)-(F2-4F)=0

We add all the numbers together, and all the variables

-(+F^2-4F)+F=0

We get rid of parentheses

-F^2+4F+F=0

We add all the numbers together, and all the variables

-1F^2+5F=0

a = -1; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-1)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-1}=\frac{-10}{-2}
=+5
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-1}=\frac{0}{-2}
=0
$