5/9c+1/3c=3/10

Solution for 5/9c+1/3c=3/10 equation:

5/9c+1/3c=3/10

We move all terms to the left:

5/9c+1/3c-(3/10)=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

5/9c+1/3c-(+3/10)=0

We get rid of parentheses

5/9c+1/3c-3/10=0

We calculate fractions


(-243c^2)/270c^2+150c/270c^2+90c/270c^2=0

We multiply all the terms by the denominator


(-243c^2)+150c+90c=0

We add all the numbers together, and all the variables

(-243c^2)+240c=0

We get rid of parentheses

-243c^2+240c=0

a = -243; b = 240; c = 0;
Δ = b2-4ac
Δ = 2402-4·(-243)·0
Δ = 57600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{57600}=240$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(240)-240}{2*-243}=\frac{-480}{-486}
=80/81
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(240)+240}{2*-243}=\frac{0}{-486}
=0
$