k=3/3k=

Solution for k=3/3k= equation:

k=3/3k=

We move all terms to the left:

k-(3/3k)=0


Domain of the equation: 3k)!=0

k!=0/1

k!=0

k∈R


We add all the numbers together, and all the variables

k-(+3/3k)=0

We get rid of parentheses

k-3/3k=0

We multiply all the terms by the denominator


k*3k-3=0

Wy multiply elements

3k^2-3=0

a = 3; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·3·(-3)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*3}=\frac{-6}{6}
=-1
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*3}=\frac{6}{6}
=1
$