1/3f+40=5/7f

Solution for 1/3f+40=5/7f equation:

1/3f+40=5/7f

We move all terms to the left:

1/3f+40-(5/7f)=0


Domain of the equation: 3f!=0

f!=0/3

f!=0

f∈R



Domain of the equation: 7f)!=0

f!=0/1

f!=0

f∈R


We add all the numbers together, and all the variables

1/3f-(+5/7f)+40=0

We get rid of parentheses

1/3f-5/7f+40=0

We calculate fractions


7f/21f^2+(-15f)/21f^2+40=0

We multiply all the terms by the denominator


7f+(-15f)+40*21f^2=0

Wy multiply elements

840f^2+7f+(-15f)=0

We get rid of parentheses

840f^2+7f-15f=0

We add all the numbers together, and all the variables

840f^2-8f=0

a = 840; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·840·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*840}=\frac{0}{1680}
=0
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*840}=\frac{16}{1680}
=1/105
$