7/10c+1/5c=1/3

Solution for 7/10c+1/5c=1/3 equation:

7/10c+1/5c=1/3

We move all terms to the left:

7/10c+1/5c-(1/3)=0


Domain of the equation: 10c!=0

c!=0/10

c!=0

c∈R



Domain of the equation: 5c!=0

c!=0/5

c!=0

c∈R


We add all the numbers together, and all the variables

7/10c+1/5c-(+1/3)=0

We get rid of parentheses

7/10c+1/5c-1/3=0

We calculate fractions


(-250c^2)/450c^2+315c/450c^2+90c/450c^2=0

We multiply all the terms by the denominator


(-250c^2)+315c+90c=0

We add all the numbers together, and all the variables

(-250c^2)+405c=0

We get rid of parentheses

-250c^2+405c=0

a = -250; b = 405; c = 0;
Δ = b2-4ac
Δ = 4052-4·(-250)·0
Δ = 164025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{164025}=405$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(405)-405}{2*-250}=\frac{-810}{-500}
=1+31/50
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(405)+405}{2*-250}=\frac{0}{-500}
=0
$