3/(10f)+24=4-1/5f

Solution for 3/(10f)+24=4-1/5f equation:

3/(10f)+24=4-1/5f

We move all terms to the left:

3/(10f)+24-(4-1/5f)=0


Domain of the equation: 10f!=0

f!=0/10

f!=0

f∈R



Domain of the equation: 5f)!=0

f!=0/1

f!=0

f∈R


We add all the numbers together, and all the variables

3/10f-(-1/5f+4)+24=0

We get rid of parentheses

3/10f+1/5f-4+24=0

We calculate fractions


15f/50f^2+10f/50f^2-4+24=0

We add all the numbers together, and all the variables

15f/50f^2+10f/50f^2+20=0

We multiply all the terms by the denominator


15f+10f+20*50f^2=0

We add all the numbers together, and all the variables

25f+20*50f^2=0

Wy multiply elements

1000f^2+25f=0

a = 1000; b = 25; c = 0;
Δ = b2-4ac
Δ = 252-4·1000·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*1000}=\frac{-50}{2000}
=-1/40
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*1000}=\frac{0}{2000}
=0
$