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

We get rid of parentheses

f-4/5f+2=0

We multiply all the terms by the denominator


f*5f+2*5f-4=0

Wy multiply elements

5f^2+10f-4=0

a = 5; b = 10; c = -4;
Δ = b2-4ac
Δ = 102-4·5·(-4)
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6\sqrt{5}}{2*5}=\frac{-10-6\sqrt{5}}{10}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6\sqrt{5}}{2*5}=\frac{-10+6\sqrt{5}}{10}
$