-20k(k-3)+40K=20-2K

Solution for -20k(k-3)+40K=20-2K equation:

-20k(k-3)+40=20-2

We move all terms to the left:

-20k(k-3)+40-(20-2)=0

We add all the numbers together, and all the variables

-20k(k-3)+40-18=0

We add all the numbers together, and all the variables

-20k(k-3)+22=0

We multiply parentheses

-20k^2+60k+22=0

a = -20; b = 60; c = +22;
Δ = b2-4ac
Δ = 602-4·(-20)·22
Δ = 5360
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:

$\sqrt{\Delta}=\sqrt{5360}=\sqrt{16*335}=\sqrt{16}*\sqrt{335}=4\sqrt{335}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-4\sqrt{335}}{2*-20}=\frac{-60-4\sqrt{335}}{-40}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+4\sqrt{335}}{2*-20}=\frac{-60+4\sqrt{335}}{-40}
$