5=(7/2)(3x+2)

Solution for 5=(7/2)(3x+2) equation:

5=(7/2)(3x+2)

We move all terms to the left:

5-((7/2)(3x+2))=0


Domain of the equation: 2)(3x+2))!=0

x∈R


We add all the numbers together, and all the variables

-((+7/2)(3x+2))+5=0

We multiply parentheses ..

-((+21x^2+7/2*2))+5=0

We multiply all the terms by the denominator


-((+21x^2+7+5*2*2))=0


We calculate terms in parentheses: -((+21x^2+7+5*2*2)), so:

(+21x^2+7+5*2*2)

We get rid of parentheses

21x^2+7+5*2*2

We add all the numbers together, and all the variables

21x^2+27

Back to the equation:

-(21x^2+27)


We get rid of parentheses

-21x^2-27=0

a = -21; b = 0; c = -27;
Δ = b2-4ac
Δ = 02-4·(-21)·(-27)
Δ = -2268
Delta is less than zero, so there is no solution for the equation