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-2+4/5f-3=f
We move all terms to the left:
-2+4/5f-3-(f)=0
Domain of the equation: 5f!=0We add all the numbers together, and all the variables
f!=0/5
f!=0
f∈R
-1f+4/5f-5=0
We multiply all the terms by the denominator
-1f*5f-5*5f+4=0
Wy multiply elements
-5f^2-25f+4=0
a = -5; b = -25; c = +4;
Δ = b2-4ac
Δ = -252-4·(-5)·4
Δ = 705
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}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-\sqrt{705}}{2*-5}=\frac{25-\sqrt{705}}{-10} $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{705}}{2*-5}=\frac{25+\sqrt{705}}{-10} $
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