(2/3)(r)-(4/3)=(24/5)

Solution for (2/3)(r)-(4/3)=(24/5) equation:

(2/3)(r)-(4/3)=(24/5)

We move all terms to the left:

(2/3)(r)-(4/3)-((24/5))=0


Domain of the equation: 3)r!=0

r!=0/1

r!=0

r∈R


We add all the numbers together, and all the variables

(+2/3)r-(+4/3)-((+24/5))=0

We multiply parentheses

2r^2-(+4/3)-((+24/5))=0

We get rid of parentheses

2r^2-4/3-((+24/5))=0

We calculate fractions


2r^2+()/()+()/()=0

We add all the numbers together, and all the variables

2r^2+2=0

a = 2; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·2·2
Δ = -16
Delta is less than zero, so there is no solution for the equation