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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

-2k^2+2k+2=0

a = -2; b = 2; c = +2;
Δ = b2-4ac
Δ = 22-4·(-2)·2
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{5}}{2*-2}=\frac{-2-2\sqrt{5}}{-4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{5}}{2*-2}=\frac{-2+2\sqrt{5}}{-4}
$