(m/2)(m+3)=(m+2)(m-2)

Solution for (m/2)(m+3)=(m+2)(m-2) equation:

(m/2)(m+3)=(m+2)(m-2)

We move all terms to the left:

(m/2)(m+3)-((m+2)(m-2))=0


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

m∈R


We add all the numbers together, and all the variables

(+m/2)(m+3)-((m+2)(m-2))=0

We use the square of the difference formula

m^2+(+m/2)(m+3)+4=0

We multiply parentheses ..

m^2+(+m^2+3m)+4=0

We get rid of parentheses

m^2+m^2+3m+4=0

We add all the numbers together, and all the variables

2m^2+3m+4=0

a = 2; b = 3; c = +4;
Δ = b2-4ac
Δ = 32-4·2·4
Δ = -23
Delta is less than zero, so there is no solution for the equation