2(-4/2+m)+3=4(m-3)+51/2

Solution for 2(-4/2+m)+3=4(m-3)+51/2 equation:

2(-4/2+m)+3=4(m-3)+51/2

We move all terms to the left:

2(-4/2+m)+3-(4(m-3)+51/2)=0


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

We move all terms containing m to the left, all other terms to the right

m)!=-2

m!=-2/1

m!=-2

m∈R


We add all the numbers together, and all the variables

2(m-2)-(4(m-3)+51/2)+3=0

We multiply parentheses

2m-(4(m-3)+51/2)-4+3=0

We multiply all the terms by the denominator


2m*2)-(4(m-3)+51-4*2)+3*2)=0

We add all the numbers together, and all the variables

2m*2)-(4(m-3)=0

Wy multiply elements

4m^2=0

a = 4; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·4·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$m=\frac{-b}{2a}=\frac{0}{8}=0$