9/10c+1/5c=2/3

Solution for 9/10c+1/5c=2/3 equation:

9/10c+1/5c=2/3

We move all terms to the left:

9/10c+1/5c-(2/3)=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

9/10c+1/5c-(+2/3)=0

We get rid of parentheses

9/10c+1/5c-2/3=0

We calculate fractions


(-500c^2)/450c^2+405c/450c^2+90c/450c^2=0

We multiply all the terms by the denominator


(-500c^2)+405c+90c=0

We add all the numbers together, and all the variables

(-500c^2)+495c=0

We get rid of parentheses

-500c^2+495c=0

a = -500; b = 495; c = 0;
Δ = b2-4ac
Δ = 4952-4·(-500)·0
Δ = 245025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{245025}=495$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(495)-495}{2*-500}=\frac{-990}{-1000}
=99/100
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(495)+495}{2*-500}=\frac{0}{-1000}
=0
$