3/5f+24=4-5/1f

Solution for 3/5f+24=4-5/1f equation:

3/5f+24=4-5/1f

We move all terms to the left:

3/5f+24-(4-5/1f)=0


Domain of the equation: 5f!=0

f!=0/5

f!=0

f∈R



Domain of the equation: 1f)!=0

f!=0/1

f!=0

f∈R


We add all the numbers together, and all the variables

3/5f-(-5/1f+4)+24=0

We get rid of parentheses

3/5f+5/1f-4+24=0

We calculate fractions


3f/5f^2+25f/5f^2-4+24=0

We add all the numbers together, and all the variables

3f/5f^2+25f/5f^2+20=0

We multiply all the terms by the denominator


3f+25f+20*5f^2=0

We add all the numbers together, and all the variables

28f+20*5f^2=0

Wy multiply elements

100f^2+28f=0

a = 100; b = 28; c = 0;
Δ = b2-4ac
Δ = 282-4·100·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-28}{2*100}=\frac{-56}{200}
=-7/25
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+28}{2*100}=\frac{0}{200}
=0
$