2k^2-32k+16=0

Solution for 2k^2-32k+16=0 equation:

2k^2-32k+16=0

a = 2; b = -32; c = +16;
Δ = b2-4ac
Δ = -322-4·2·16
Δ = 896
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:

$\sqrt{\Delta}=\sqrt{896}=\sqrt{64*14}=\sqrt{64}*\sqrt{14}=8\sqrt{14}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{14}}{2*2}=\frac{32-8\sqrt{14}}{4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{14}}{2*2}=\frac{32+8\sqrt{14}}{4}
$