1k^2-10k+20=0

Solution for 1k^2-10k+20=0 equation:

1k^2-10k+20=0

We add all the numbers together, and all the variables

k^2-10k+20=0

a = 1; b = -10; c = +20;
Δ = b2-4ac
Δ = -102-4·1·20
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{-(-10)-2\sqrt{5}}{2*1}=\frac{10-2\sqrt{5}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{5}}{2*1}=\frac{10+2\sqrt{5}}{2}
$