12=(7/5)(5x+2)

Solution for 12=(7/5)(5x+2) equation:

12=(7/5)(5x+2)

We move all terms to the left:

12-((7/5)(5x+2))=0


Domain of the equation: 5)(5x+2))!=0

x∈R


We add all the numbers together, and all the variables

-((+7/5)(5x+2))+12=0

We multiply parentheses ..

-((+35x^2+7/5*2))+12=0

We multiply all the terms by the denominator


-((+35x^2+7+12*5*2))=0


We calculate terms in parentheses: -((+35x^2+7+12*5*2)), so:

(+35x^2+7+12*5*2)

We get rid of parentheses

35x^2+7+12*5*2

We add all the numbers together, and all the variables

35x^2+127

Back to the equation:

-(35x^2+127)


We get rid of parentheses

-35x^2-127=0

a = -35; b = 0; c = -127;
Δ = b2-4ac
Δ = 02-4·(-35)·(-127)
Δ = -17780
Delta is less than zero, so there is no solution for the equation