F(x)=2+2/5x

Solution for F(x)=2+2/5x equation:

(F)=2+2/5F

We move all terms to the left:

(F)-(2+2/5F)=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-(2/5F+2)=0

We get rid of parentheses

F-2/5F-2=0

We multiply all the terms by the denominator


F*5F-2*5F-2=0

Wy multiply elements

5F^2-10F-2=0

a = 5; b = -10; c = -2;
Δ = b2-4ac
Δ = -102-4·5·(-2)
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{35}}{2*5}=\frac{10-2\sqrt{35}}{10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{35}}{2*5}=\frac{10+2\sqrt{35}}{10}
$