3-3k=3k(2k-1)

Solution for 3-3k=3k(2k-1) equation:

3-3k=3k(2k-1)

We move all terms to the left:

3-3k-(3k(2k-1))=0


We calculate terms in parentheses: -(3k(2k-1)), so:

3k(2k-1)

We multiply parentheses

6k^2-3k

Back to the equation:

-(6k^2-3k)


We get rid of parentheses

-6k^2-3k+3k+3=0

We add all the numbers together, and all the variables

-6k^2+3=0

a = -6; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-6)·3
Δ = 72
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{2}}{2*-6}=\frac{0-6\sqrt{2}}{-12}
=-\frac{6\sqrt{2}}{-12}
=-\frac{\sqrt{2}}{-2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{2}}{2*-6}=\frac{0+6\sqrt{2}}{-12}
=\frac{6\sqrt{2}}{-12}
=\frac{\sqrt{2}}{-2}
$