(2/3)(b+3)=(1/2)+b+1

Solution for (2/3)(b+3)=(1/2)+b+1 equation:

(2/3)(b+3)=(1/2)+b+1

We move all terms to the left:

(2/3)(b+3)-((1/2)+b+1)=0


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

b∈R



Domain of the equation: 2)+b+1)!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

(+2/3)(b+3)-((+1/2)+b+1)=0

We multiply parentheses ..

(+2b^2+2/3*3)-((+1/2)+b+1)=0

We calculate fractions


(2b^2+b+4)/18b^2+()/18b^2=0

We multiply all the terms by the denominator


(2b^2+b+4)+()=0

We add all the numbers together, and all the variables

(2b^2+b+4)=0

We get rid of parentheses

2b^2+b+4=0

a = 2; b = 1; c = +4;
Δ = b2-4ac
Δ = 12-4·2·4
Δ = -31
Delta is less than zero, so there is no solution for the equation