F=9/5c+32;c

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



=9/5F+32F

We move all terms to the left:

-(9/5F+32F)=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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