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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We calculate fractions


(-567c^2)/1728c^2+1344c/1728c^2+576c/1728c^2=0

We multiply all the terms by the denominator


(-567c^2)+1344c+576c=0

We add all the numbers together, and all the variables

(-567c^2)+1920c=0

We get rid of parentheses

-567c^2+1920c=0

a = -567; b = 1920; c = 0;
Δ = b2-4ac
Δ = 19202-4·(-567)·0
Δ = 3686400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3686400}=1920$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1920)-1920}{2*-567}=\frac{-3840}{-1134}
=3+73/189
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1920)+1920}{2*-567}=\frac{0}{-1134}
=0
$