-2k^2+16k-5=0

Solution for -2k^2+16k-5=0 equation:

-2k^2+16k-5=0

a = -2; b = 16; c = -5;
Δ = b2-4ac
Δ = 162-4·(-2)·(-5)
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-6\sqrt{6}}{2*-2}=\frac{-16-6\sqrt{6}}{-4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+6\sqrt{6}}{2*-2}=\frac{-16+6\sqrt{6}}{-4}
$