F=9/5c+32(c)

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Solution for F=9/5c+32(c) equation:



=9/5F+32(F)
We move all terms to the left:
-(9/5F+32(F))=0
Domain of the equation: 5F+32F)!=0
F∈R
We add all the numbers together, and all the variables
-(+32F+9/5F)=0
We get rid of parentheses
-32F-9/5F=0
We multiply all the terms by the denominator
-32F*5F-9=0
Wy multiply elements
-160F^2-9=0
a = -160; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·(-160)·(-9)
Δ = -5760
Delta is less than zero, so there is no solution for the equation

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