1/9c+1/3c=1/4

Solution for 1/9c+1/3c=1/4 equation:

1/9c+1/3c=1/4

We move all terms to the left:

1/9c+1/3c-(1/4)=0


Domain of the equation: 9c!=0

c!=0/9

c!=0

c∈R



Domain of the equation: 3c!=0

c!=0/3

c!=0

c∈R


We add all the numbers together, and all the variables

1/9c+1/3c-(+1/4)=0

We get rid of parentheses

1/9c+1/3c-1/4=0

We calculate fractions


(-81c^2)/432c^2+48c/432c^2+144c/432c^2=0

We multiply all the terms by the denominator


(-81c^2)+48c+144c=0

We add all the numbers together, and all the variables

(-81c^2)+192c=0

We get rid of parentheses

-81c^2+192c=0

a = -81; b = 192; c = 0;
Δ = b2-4ac
Δ = 1922-4·(-81)·0
Δ = 36864
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36864}=192$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(192)-192}{2*-81}=\frac{-384}{-162}
=2+10/27
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(192)+192}{2*-81}=\frac{0}{-162}
=0
$