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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)!=0We add all the numbers together, and all the variables
f!=0/1
f!=0
f∈R
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 $
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