8/9c+1/3c=7/10

Solution for 8/9c+1/3c=7/10 equation:

8/9c+1/3c=7/10

We move all terms to the left:

8/9c+1/3c-(7/10)=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

8/9c+1/3c-(+7/10)=0

We get rid of parentheses

8/9c+1/3c-7/10=0

We calculate fractions


(-567c^2)/270c^2+240c/270c^2+90c/270c^2=0

We multiply all the terms by the denominator


(-567c^2)+240c+90c=0

We add all the numbers together, and all the variables

(-567c^2)+330c=0

We get rid of parentheses

-567c^2+330c=0

a = -567; b = 330; c = 0;
Δ = b2-4ac
Δ = 3302-4·(-567)·0
Δ = 108900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{108900}=330$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(330)-330}{2*-567}=\frac{-660}{-1134}
=110/189
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(330)+330}{2*-567}=\frac{0}{-1134}
=0
$