9/5c+32=c

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

9/5c+32=c

We move all terms to the left:

9/5c+32-(c)=0


Domain of the equation: 5c!=0

c!=0/5

c!=0

c∈R


We add all the numbers together, and all the variables

-1c+9/5c+32=0

We multiply all the terms by the denominator


-1c*5c+32*5c+9=0

Wy multiply elements

-5c^2+160c+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:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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