(-k+1,3)(3,k+3)=3

Solution for (-k+1,3)(3,k+3)=3 equation:

(-k+1.3)(3.k+3)=3

We move all terms to the left:

(-k+1.3)(3.k+3)-(3)=0

We add all the numbers together, and all the variables

(-1k+1.3)(3.k+3)-3=0

We multiply parentheses ..

(-3k^2-3k+3.9k+3.9)-3=0

We get rid of parentheses

-3k^2-3k+3.9k+3.9-3=0

We add all the numbers together, and all the variables

-3k^2+0.9k+0.9=0

a = -3; b = 0.9; c = +0.9;
Δ = b2-4ac
Δ = 0.92-4·(-3)·0.9
Δ = 11.61
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.9)-\sqrt{11.61}}{2*-3}=\frac{-0.9-\sqrt{11.61}}{-6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.9)+\sqrt{11.61}}{2*-3}=\frac{-0.9+\sqrt{11.61}}{-6}
$