2f(5f-2)-10(f^2-3f+6)=-8f(f+4)+4(2f^2+5)

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Solution for 2f(5f-2)-10(f^2-3f+6)=-8f(f+4)+4(2f^2+5) equation:



2f(5f-2)-10(f^2-3f+6)=-8f(f+4)+4(2f^2+5)
We move all terms to the left:
2f(5f-2)-10(f^2-3f+6)-(-8f(f+4)+4(2f^2+5))=0
We multiply parentheses
10f^2-10f^2-4f+30f-(-8f(f+4)+4(2f^2+5))-60=0
We calculate terms in parentheses: -(-8f(f+4)+4(2f^2+5)), so:
-8f(f+4)+4(2f^2+5)
We multiply parentheses
-8f^2+8f^2-32f+20
We add all the numbers together, and all the variables
-32f+20
Back to the equation:
-(-32f+20)
We add all the numbers together, and all the variables
26f-(-32f+20)-60=0
We get rid of parentheses
26f+32f-20-60=0
We add all the numbers together, and all the variables
58f-80=0
We move all terms containing f to the left, all other terms to the right
58f=80
f=80/58
f=1+11/29

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