F(x)=-16+2/5x

Solution for F(x)=-16+2/5x equation:

(F)=-16+2/5F

We move all terms to the left:

(F)-(-16+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-16)=0

We get rid of parentheses

F-2/5F+16=0

We multiply all the terms by the denominator


F*5F+16*5F-2=0

Wy multiply elements

5F^2+80F-2=0

a = 5; b = 80; c = -2;
Δ = b2-4ac
Δ = 802-4·5·(-2)
Δ = 6440
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:

$\sqrt{\Delta}=\sqrt{6440}=\sqrt{4*1610}=\sqrt{4}*\sqrt{1610}=2\sqrt{1610}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-2\sqrt{1610}}{2*5}=\frac{-80-2\sqrt{1610}}{10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+2\sqrt{1610}}{2*5}=\frac{-80+2\sqrt{1610}}{10}
$