3f+3/5f=4/5f+2f+4

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

3f+3/5f=4/5f+2f+4

We move all terms to the left:

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


Domain of the equation: 5f!=0

f!=0/5

f!=0

f∈R



Domain of the equation: 5f+2f+4)!=0

f∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


3f*5f-2f*5f-4*5f+3-4=0

We add all the numbers together, and all the variables

3f*5f-2f*5f-4*5f-1=0

Wy multiply elements

15f^2-10f^2-20f-1=0

We add all the numbers together, and all the variables

5f^2-20f-1=0

a = 5; b = -20; c = -1;
Δ = b2-4ac
Δ = -202-4·5·(-1)
Δ = 420
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:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{105}}{2*5}=\frac{20-2\sqrt{105}}{10}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{105}}{2*5}=\frac{20+2\sqrt{105}}{10}
$