(1/8)m^2+4m+12=0

Solution for (1/8)m^2+4m+12=0 equation:

(1/8)m^2+4m+12=0


Domain of the equation: 8)m^2!=0

m!=0/1

m!=0

m∈R


We add all the numbers together, and all the variables

(+1/8)m^2+4m+12=0

We add all the numbers together, and all the variables

4m+(+1/8)m^2+12=0

We multiply all the terms by the denominator


4m*8)m^2+(+12*8)m^2+1=0

We add all the numbers together, and all the variables

4m*8)m^2+96m^2+1=0

Wy multiply elements

32m^2=0

a = 32; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·32·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$m=\frac{-b}{2a}=\frac{0}{64}=0$