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

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

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

We move all terms to the left:

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


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

F!=0/1

F!=0

F∈R


We multiply all the terms by the denominator


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

Wy multiply elements

5F^2-2=0

a = 5; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·5·(-2)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{10}}{2*5}=\frac{0-2\sqrt{10}}{10}
=-\frac{2\sqrt{10}}{10}
=-\frac{\sqrt{10}}{5}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{10}}{2*5}=\frac{0+2\sqrt{10}}{10}
=\frac{2\sqrt{10}}{10}
=\frac{\sqrt{10}}{5}
$