(1/2k)(4k)+12=2k^2+12

Solution for (1/2k)(4k)+12=2k^2+12 equation:

(1/2k)(4k)+12=2k^2+12

We move all terms to the left:

(1/2k)(4k)+12-(2k^2+12)=0


Domain of the equation: 2k)4k!=0

k!=0/1

k!=0

k∈R


We add all the numbers together, and all the variables

(+1/2k)4k-(2k^2+12)+12=0

We multiply parentheses

4k^2-(2k^2+12)+12=0

We get rid of parentheses

4k^2-2k^2-12+12=0

We add all the numbers together, and all the variables

2k^2=0

a = 2; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·2·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$k=\frac{-b}{2a}=\frac{0}{4}=0$