1/6c+5/2c=7/3

Solution for 1/6c+5/2c=7/3 equation:

1/6c+5/2c=7/3

We move all terms to the left:

1/6c+5/2c-(7/3)=0


Domain of the equation: 6c!=0

c!=0/6

c!=0

c∈R



Domain of the equation: 2c!=0

c!=0/2

c!=0

c∈R


We add all the numbers together, and all the variables

1/6c+5/2c-(+7/3)=0

We get rid of parentheses

1/6c+5/2c-7/3=0

We calculate fractions


(-168c^2)/108c^2+18c/108c^2+270c/108c^2=0

We multiply all the terms by the denominator


(-168c^2)+18c+270c=0

We add all the numbers together, and all the variables

(-168c^2)+288c=0

We get rid of parentheses

-168c^2+288c=0

a = -168; b = 288; c = 0;
Δ = b2-4ac
Δ = 2882-4·(-168)·0
Δ = 82944
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{82944}=288$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(288)-288}{2*-168}=\frac{-576}{-336}
=1+5/7
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(288)+288}{2*-168}=\frac{0}{-336}
=0
$