f=-2+4/5f-3

Solution for f=-2+4/5f-3 equation:

f=-2+4/5f-3

We move all terms to the left:

f-(-2+4/5f-3)=0


Domain of the equation: 5f-3)!=0

f∈R


We add all the numbers together, and all the variables

f-(4/5f-5)=0

We get rid of parentheses

f-4/5f+5=0

We multiply all the terms by the denominator


f*5f+5*5f-4=0

Wy multiply elements

5f^2+25f-4=0

a = 5; b = 25; c = -4;
Δ = b2-4ac
Δ = 252-4·5·(-4)
Δ = 705
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-\sqrt{705}}{2*5}=\frac{-25-\sqrt{705}}{10}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{705}}{2*5}=\frac{-25+\sqrt{705}}{10}
$