F(x)=2,8x(1-x)

Solution for F(x)=2,8x(1-x) equation:

(F)=2.8F(1-F)

We move all terms to the left:

(F)-(2.8F(1-F))=0

We add all the numbers together, and all the variables

F-(2.8F(-1F+1))=0


We calculate terms in parentheses: -(2.8F(-1F+1)), so:

2.8F(-1F+1)

We multiply parentheses

-2F^2+2F

Back to the equation:

-(-2F^2+2F)


We get rid of parentheses

2F^2-2F+F=0

We add all the numbers together, and all the variables

2F^2-1F=0

a = 2; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·2·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*2}=\frac{0}{4}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*2}=\frac{2}{4}
=1/2
$