28+20k-k^2=0

Solution for 28+20k-k^2=0 equation:

28+20k-k^2=0

We add all the numbers together, and all the variables

-1k^2+20k+28=0

a = -1; b = 20; c = +28;
Δ = b2-4ac
Δ = 202-4·(-1)·28
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16\sqrt{2}}{2*-1}=\frac{-20-16\sqrt{2}}{-2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16\sqrt{2}}{2*-1}=\frac{-20+16\sqrt{2}}{-2}
$