(9/5)c+32=68

Solution for (9/5)c+32=68 equation:

(9/5)c+32=68

We move all terms to the left:

(9/5)c+32-(68)=0


Domain of the equation: 5)c!=0

c!=0/1

c!=0

c∈R


We add all the numbers together, and all the variables

(+9/5)c+32-68=0

We add all the numbers together, and all the variables

(+9/5)c-36=0

We multiply parentheses

9c^2-36=0

a = 9; b = 0; c = -36;
Δ = b2-4ac
Δ = 02-4·9·(-36)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36}{2*9}=\frac{-36}{18}
=-2
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36}{2*9}=\frac{36}{18}
=2
$