0=k^2-12k+3k

Solution for 0=k^2-12k+3k equation:

0=k^2-12k+3k

We move all terms to the left:

0-(k^2-12k+3k)=0

We add all the numbers together, and all the variables

-(k^2-12k+3k)=0

We get rid of parentheses

-k^2+12k-3k=0

We add all the numbers together, and all the variables

-1k^2+9k=0

a = -1; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·(-1)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*-1}=\frac{-18}{-2}
=+9
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*-1}=\frac{0}{-2}
=0
$