2k^2-8k-3=0

Solution for 2k^2-8k-3=0 equation:

2k^2-8k-3=0

a = 2; b = -8; c = -3;
Δ = b2-4ac
Δ = -82-4·2·(-3)
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{22}}{2*2}=\frac{8-2\sqrt{22}}{4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{22}}{2*2}=\frac{8+2\sqrt{22}}{4}
$