((1/2)c+10)=(5-7c)/6

Solution for ((1/2)c+10)=(5-7c)/6 equation:

((1/2)c+10)=(5-7c)/6

We move all terms to the left:

((1/2)c+10)-((5-7c)/6)=0


Domain of the equation: 2)c+10)!=0

c!=0/1

c!=0

c∈R


We add all the numbers together, and all the variables

((+1/2)c+10)-((-7c+5)/6)=0

We calculate fractions


()/2c-7c+5)*2)c)/2c+10)*6))+10)*6))+(-((=0

We add all the numbers together, and all the variables

-7c+()/2c+5)*2)c)/2c+10)*6))+10)*6))+(-((=0

We calculate fractions


-7c+(()*2c)/4c^2+5*2)c)*2c)/4c^2=0

We multiply all the terms by the denominator


-7c*4c^2+(()*2c)+5*2)c)*2c)=0


We calculate terms in parentheses: +(()*2c), so:

()*2c


Wy multiply elements

-28c^3+20c^2+(()*2c)=0


We calculate terms in parentheses: +(()*2c), so:

()*2c


We do not support ecpression: c^3