-10k^2+20k=0

Solution for -10k^2+20k=0 equation:

-10k^2+20k=0

a = -10; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·(-10)·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*-10}=\frac{-40}{-20}
=+2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*-10}=\frac{0}{-20}
=0
$