F(x)=2/5x+4

Solution for F(x)=2/5x+4 equation:

(F)=2/5F+4

We move all terms to the left:

(F)-(2/5F+4)=0


Domain of the equation: 5F+4)!=0

F∈R


We get rid of parentheses

F-2/5F-4=0

We multiply all the terms by the denominator


F*5F-4*5F-2=0

Wy multiply elements

5F^2-20F-2=0

a = 5; b = -20; c = -2;
Δ = b2-4ac
Δ = -202-4·5·(-2)
Δ = 440
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:

$\sqrt{\Delta}=\sqrt{440}=\sqrt{4*110}=\sqrt{4}*\sqrt{110}=2\sqrt{110}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{110}}{2*5}=\frac{20-2\sqrt{110}}{10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{110}}{2*5}=\frac{20+2\sqrt{110}}{10}
$