5=(2/7)(10x+35)

Solution for 5=(2/7)(10x+35) equation:

5=(2/7)(10x+35)

We move all terms to the left:

5-((2/7)(10x+35))=0


Domain of the equation: 7)(10x+35))!=0

x∈R


We add all the numbers together, and all the variables

-((+2/7)(10x+35))+5=0

We multiply parentheses ..

-((+20x^2+2/7*35))+5=0

We multiply all the terms by the denominator


-((+20x^2+2+5*7*35))=0


We calculate terms in parentheses: -((+20x^2+2+5*7*35)), so:

(+20x^2+2+5*7*35)

We get rid of parentheses

20x^2+2+5*7*35

We add all the numbers together, and all the variables

20x^2+1227

Back to the equation:

-(20x^2+1227)


We get rid of parentheses

-20x^2-1227=0

a = -20; b = 0; c = -1227;
Δ = b2-4ac
Δ = 02-4·(-20)·(-1227)
Δ = -98160
Delta is less than zero, so there is no solution for the equation