5/6(12n-18)+7=3/7(21n-35)

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Solution for 5/6(12n-18)+7=3/7(21n-35) equation: 



5/6(12n-18)+7=3/7(21n-35)

We move all terms to the left:

5/6(12n-18)+7-(3/7(21n-35))=0



Domain of the equation: 6(12n-18)!=0

n∈R





Domain of the equation: 7(21n-35))!=0

n∈R



We calculate fractions


(35n2/(6(12n-18)*7(21n-35)))+(-18n1/(6(12n-18)*7(21n-35)))+7=0



We calculate terms in parentheses: +(35n2/(6(12n-18)*7(21n-35))), so:

35n2/(6(12n-18)*7(21n-35))

We multiply all the terms by the denominator


35n2

We add all the numbers together, and all the variables

35n^2

Back to the equation:

+(35n^2)





We calculate terms in parentheses: +(-18n1/(6(12n-18)*7(21n-35))), so:

-18n1/(6(12n-18)*7(21n-35))

We multiply all the terms by the denominator


-18n1

We add all the numbers together, and all the variables

-18n

Back to the equation:

+(-18n)



We get rid of parentheses

35n^2-18n+7=0

a = 35; b = -18; c = +7;
Δ = b2-4ac
Δ = -182-4·35·7
Δ = -656
Delta is less than zero, so there is no solution for the equation

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