7/9c+1/3c=1/5

Solution for 7/9c+1/3c=1/5 equation:

7/9c+1/3c=1/5

We move all terms to the left:

7/9c+1/3c-(1/5)=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

7/9c+1/3c-(+1/5)=0

We get rid of parentheses

7/9c+1/3c-1/5=0

We calculate fractions


(-81c^2)/675c^2+525c/675c^2+225c/675c^2=0

We multiply all the terms by the denominator


(-81c^2)+525c+225c=0

We add all the numbers together, and all the variables

(-81c^2)+750c=0

We get rid of parentheses

-81c^2+750c=0

a = -81; b = 750; c = 0;
Δ = b2-4ac
Δ = 7502-4·(-81)·0
Δ = 562500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{562500}=750$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(750)-750}{2*-81}=\frac{-1500}{-162}
=9+7/27
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(750)+750}{2*-81}=\frac{0}{-162}
=0
$