1c=9/5c+32

Solution for 1c=9/5c+32 equation:

1c=9/5c+32

We move all terms to the left:

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


Domain of the equation: 5c+32)!=0

c∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

c-9/5c-32=0

We multiply all the terms by the denominator


c*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}
$