F(x)=400-4/5x

Solution for F(x)=400-4/5x equation:

(F)=400-4/5F

We move all terms to the left:

(F)-(400-4/5F)=0


Domain of the equation: 5F)!=0

F!=0/1

F!=0

F∈R


We add all the numbers together, and all the variables

F-(-4/5F+400)=0

We get rid of parentheses

F+4/5F-400=0

We multiply all the terms by the denominator


F*5F-400*5F+4=0

Wy multiply elements

5F^2-2000F+4=0

a = 5; b = -2000; c = +4;
Δ = b2-4ac
Δ = -20002-4·5·4
Δ = 3999920
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:

$\sqrt{\Delta}=\sqrt{3999920}=\sqrt{16*249995}=\sqrt{16}*\sqrt{249995}=4\sqrt{249995}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2000)-4\sqrt{249995}}{2*5}=\frac{2000-4\sqrt{249995}}{10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2000)+4\sqrt{249995}}{2*5}=\frac{2000+4\sqrt{249995}}{10}
$