(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

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+2m+6)+4=0

We get rid of parentheses

m^2+m^2+3m+2m+6+4=0

We add all the numbers together, and all the variables

2m^2+5m+10=0

a = 2; b = 5; c = +10;
Δ = b2-4ac
Δ = 52-4·2·10
Δ = -55
Delta is less than zero, so there is no solution for the equation