(5/9)(f-32)=-20

Solution for (5/9)(f-32)=-20 equation:

(5/9)(f-32)=-20

We move all terms to the left:

(5/9)(f-32)-(-20)=0


Domain of the equation: 9)(f-32)!=0

f∈R


We add all the numbers together, and all the variables

(+5/9)(f-32)-(-20)=0

We add all the numbers together, and all the variables

(+5/9)(f-32)+20=0

We multiply parentheses ..

(+5f^2+5/9*-32)+20=0

We multiply all the terms by the denominator


(+5f^2+5+20*9*-32)=0

We get rid of parentheses

5f^2+5-32+20*9*=0

We add all the numbers together, and all the variables

5f^2=0

a = 5; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·5·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$f=\frac{-b}{2a}=\frac{0}{10}=0$