k^2-20k+16=0

Solution for k^2-20k+16=0 equation:

k^2-20k+16=0

a = 1; b = -20; c = +16;
Δ = b2-4ac
Δ = -202-4·1·16
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{21}}{2*1}=\frac{20-4\sqrt{21}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{21}}{2*1}=\frac{20+4\sqrt{21}}{2}
$