f=9/5f-32

Solution for f=9/5f-32 equation:

f=9/5f-32

We move all terms to the left:

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


Domain of the equation: 5f-32)!=0

f∈R


We get rid of parentheses

f-9/5f+32=0

We multiply all the terms by the denominator


f*5f+32*5f-9=0

Wy multiply elements

5f^2+160f-9=0

a = 5; b = 160; c = -9;
Δ = b2-4ac
Δ = 1602-4·5·(-9)
Δ = 25780
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{25780}=\sqrt{4*6445}=\sqrt{4}*\sqrt{6445}=2\sqrt{6445}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-2\sqrt{6445}}{2*5}=\frac{-160-2\sqrt{6445}}{10}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+2\sqrt{6445}}{2*5}=\frac{-160+2\sqrt{6445}}{10}
$