F(c)=9/5c+32

Solution for F(c)=9/5c+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}
$