F(x)=4/5x+20

Solution for F(x)=4/5x+20 equation:

(F)=4/5F+20

We move all terms to the left:

(F)-(4/5F+20)=0


Domain of the equation: 5F+20)!=0

F∈R


We get rid of parentheses

F-4/5F-20=0

We multiply all the terms by the denominator


F*5F-20*5F-4=0

Wy multiply elements

5F^2-100F-4=0

a = 5; b = -100; c = -4;
Δ = b2-4ac
Δ = -1002-4·5·(-4)
Δ = 10080
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:

$\sqrt{\Delta}=\sqrt{10080}=\sqrt{144*70}=\sqrt{144}*\sqrt{70}=12\sqrt{70}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-12\sqrt{70}}{2*5}=\frac{100-12\sqrt{70}}{10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+12\sqrt{70}}{2*5}=\frac{100+12\sqrt{70}}{10}
$