2c=9/5c+32

Solution for 2c=9/5c+32 equation:

2c=9/5c+32

We move all terms to the left:

2c-(9/5c+32)=0


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

c∈R


We get rid of parentheses

2c-9/5c-32=0

We multiply all the terms by the denominator


2c*5c-32*5c-9=0

Wy multiply elements

10c^2-160c-9=0

a = 10; b = -160; c = -9;
Δ = b2-4ac
Δ = -1602-4·10·(-9)
Δ = 25960
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:

$\sqrt{\Delta}=\sqrt{25960}=\sqrt{4*6490}=\sqrt{4}*\sqrt{6490}=2\sqrt{6490}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-2\sqrt{6490}}{2*10}=\frac{160-2\sqrt{6490}}{20}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+2\sqrt{6490}}{2*10}=\frac{160+2\sqrt{6490}}{20}
$